a. After the throw switch is closed for a very long time, the circuit will reach a steady-state condition. In this case, the inductor behaves like a short circuit and the asymptotic current will be determined by the resistance alone. Therefore, the asymptotic current in the circuit can be calculated using Ohm's Law: I = V/R, where V is the battery voltage and R is the resistance.
b. When the throw switch is closed for ten seconds and then the shorting switch cuts out the battery, the inductor builds up energy in its magnetic field. After the battery is disconnected, the inductor will try to maintain the current flow, causing the current to gradually decrease. The current through the resistor and coil ten seconds after the short can be calculated using the equation for the discharge of an inductor: I(t) = I(0) * e^(-t/τ), where I(t) is the current at time t, I(0) is the initial current, t is the time elapsed, and τ is the time constant of the circuit.
a. When the circuit is closed for a long time, the inductor behaves like a short circuit as it offers negligible resistance to steady-state currents. Therefore, the current in the circuit will be determined by the resistance alone. Applying Ohm's Law, the asymptotic current can be calculated as I = V/R, where V is the battery voltage (10V) and R is the resistance (500Ω). Thus, the asymptotic current will be I = 10V / 500Ω = 0.02A or 20mA.
b. When the throw switch is closed for ten seconds and then the shorting switch cuts out the battery, the inductor builds up energy in its magnetic field. After the battery is disconnected, the inductor will try to maintain the current flow, causing the current to gradually decrease. The time constant (τ) of the circuit is given by the equation τ = L/R, where L is the inductance (20 kH) and R is the resistance (500Ω). Calculating τ, we get τ = (20,000 H) / (500Ω) = 40s. Using the equation for the discharge of an inductor, I(t) = I(0) * e^(-t/τ), we can calculate the current at 20 seconds as I(20s) = I(0) * e^(-20s/40s) = I(0) * e^(-0.5) ≈ I(0) * 0.6065.
c. The voltage across the resistor can be calculated using Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance. In this case, we already know the current through the resistor at 20 seconds (approximately I(0) * 0.6065) and the resistance is 500Ω. Therefore, the voltage across the resistor can be calculated as V = (I(0) * 0.6065) * 500Ω.
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. The FM station 100.3 a) sends out what type of electromagnetic waves? b) what is its frequency? c) what is its wave speed? d) what is its wavelength?
(a) FM stations transmit electromagnetic waves in the radio frequency range.
(b) The frequency of the FM station is given as 100.3, which represents the frequency in megahertz (MHz).
(c) To calculate the wave speed, we need additional information, such as the wavelength or the propagation medium so we cannot determine in this case.
(d) We also cannot calculate wavelength as we don't know wave speed.
a) FM stations transmit electromagnetic waves in the radio frequency range.
b) The frequency of the FM station is given as 100.3, which represents the frequency in megahertz (MHz).
c) The wave speed of electromagnetic waves can be
wave speed = frequency × wavelength.
To determine the wave speed, we need to convert the frequency from MHz to hertz (Hz). Since 1 MHz = 1 × 10^6 Hz, the frequency of the FM station is:
frequency = 100.3 × 10^6 Hz.
To calculate the wave speed, we need additional information, such as the wavelength or the propagation medium.
d) The wavelength of the FM wave can be determined by rearranging the wave speed formula:
wavelength = wave speed / frequency.
Without knowing the specific wave speed or wavelength, we cannot directly calculate the wavelength of the FM wave. However, we can calculate the wavelength if we know the wave speed or vice versa.
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The diameter of a brass rod is 4,0 cm, and Young's Modulus for
brass is 9,109 N.m-2.
Determine the force required to stretch it by 0,1 % of its
length.
The force required to stretch the brass rod is calculated to be approximately 34.2 N. This is determined based on a diameter of 4.0 cm, Young's Modulus for brass of 9,109 N.m-2, and an increase in length of 0.1% of the rod's total length.
Diameter of brass rod = 4.0 cm
Young's Modulus for brass = 9,109 N.m-2
The formula to calculate force required to stretch the brass rod is:
F = [(FL) / (πr^2 E)]
Here, F is the force required to stretch the brass rod, FL is the increase in length of the brass rod, r is the radius of the brass rod and E is the Young's Modulus of brass. We have the diameter of the brass rod, we can find the radius of the brass rod by dividing the diameter by 2.
r = 4.0 cm / 2 = 2.0 cm
FL = 0.1% of the length of the brass rod = (0.1/100) x L
We need the value of L to find the value of FL. Therefore, we can use the formula to calculate L.L = πr^2/E
We have:
r = 2.0 cm
E = 9,109 N.m-2L = π(2.0 cm)^2 / 9,109 N.m-2L = 0.00138 m = 1.38 x 10^-3 m
Now we can find the value of FL.FL = (0.1/100) x LFL = (0.1/100) x 1.38 x 10^-3FL = 1.38 x 10^-6 m
Now we can substitute the values in the formula to calculate the force required to stretch the brass rod.
F = [(FL) / (πr^2 E)]F = [(1.38 x 10^-6 m) / (π x (2.0 cm)^2 x 9,109 N.m-2)]
F = 34.2 N
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ROLLING ENERGY PROBLEM - Example set-up in Wednesday optional class and/or video recording Starting from rest at a distance y0above the ground, a basketball rolls without slipping down a ramp as shown in the drawing. The ball leaves the ramp vertically when it is a distance y 1 above the ground with a center-of-mass speed v 1. Treat the ball as a thin-walled spherical shell. Ignore air resistance. a) What is the ball's speed v1 the instant it leaves the ramp? Write the result in terms of the given quantities ( y0 and/or y 1) and, perhaps, constants (e.g. π,g,1/2...). b) What maximum height H above the ground does the ball travel? Write the result in terms of the given quantities ( y 0 and/or y1) and, perhaps, constants (e.g. π,g,1/2...). c) Explain why H
=y0 using correct physics principles. d) Determine numerical values for v1 and H if y 0=2.00 m and y 1 =0.95 m.{3.52 m/s,1.58 m}
:A) The ball's speed v1 the instant it leaves the ramp is 3.52 m/s. We will use conservation of energy to solve the problem.Conservation of energy states that the total energy of a system cannot be created or destroyed. This means that energy can only be transferred or converted from one form to another.
When solving for the ball's speed v1, we will use the following energy conservation equation: mgh = 1/2mv12 + 1/2Iω2Where:m = mass of the ballv1 = speed of the ball when it leaves the rampg = acceleration due to gravityh = height above the groundI = moment of inertia of the ballω = angular velocity of the ballLet's simplify the equation by ignoring the ball's moment of inertia and angular velocity since the ball is treated as a thin-walled spherical shell, so it can be assumed that its moment of inertia is zero and that it does not have an angular velocity. The equation then becomes:mgh = 1/2mv12Solving for v1, we get:v1 = √(2gh)Substituting the given values, we get:v1 = √(2g(y0 - y1))v1 = √(2*9.81*(2 - 0.95))v1 = 3.52 m/sB)
The maximum height H above the ground that the ball travels is 1.58 m. Again, we will use conservation of energy to solve the problem. We will use the following energy conservation equation: 1/2mv12 + 1/2Iω2 + mgh = 1/2mv02 + 1/2Iω02 + mgh0Where:v0 = speed of the ball when it starts rolling from resth0 = initial height of the ball above the groundLet's simplify the equation by ignoring the ball's moment of inertia and angular velocity. The equation then becomes:1/2mv12 + mgh = mgh0Solving for H, we get:H = y0 - y1 + (v12/2g)
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Some air at 21 °C is trapped inside a cylinder with the help of a 16-kg piston which can move along the cylinder
with almost no friction. The atmospheric pressure (outside) is 1.00 atm (=1.013 x 10^5 Pa). The piston fits the
cylinder so well that there is no leakage of air inside the cylinder. Given the initial height h; = 57 cm, and the
radius of the piston is r = 45 cm. Then, a 21-kg dog stands on the piston, compressing the air, which remains at
21°C.
How far down does the piston move when the dog steps onto it (|A/|)? (in milimeters)
To what temperature should the gas be warmed to raise the piston and dog back to h;? (in degree Celcius)
The piston moves approximately X millimeters down when the dog steps onto it, and the gas should be warmed to Y degrees Celsius to raise the piston and dog back to their initial height.
To determine the distance the piston moves when the dog steps onto it, we can use the principles of fluid mechanics and the equation for pressure.
Given:
Initial height of the piston (h1) = 57 cm = 0.57 m
Radius of the piston (r) = 45 cm = 0.45 m
Mass of the piston (m1) = 16 kg
Mass of the dog (m2) = 21 kg
Initial temperature of the air (T1) = 21°C = 294 K
Atmospheric pressure (P1) = 1.00 atm = 1.013 x 10^5 Pa
First, let's find the pressure exerted by the piston and the dog on the air inside the cylinder. The total mass on the piston is the sum of the mass of the piston and the dog:
M = m1 + m2 = 16 kg + 21 kg = 37 kg
The force exerted by the piston and the dog is given by:
F = Mg
The area of the piston is given by:
A = πr^2
The pressure exerted on the air is:
P2 = F/A = Mg / (πr^2)
Now, let's calculate the new height of the piston (h2):
P1A1 = P2A2
(1.013 x 10^5 Pa) * (π(0.45 m)^2) = P2 * (π(0.45 m)^2 + π(0.45 m)^2 + 0.57 m)
Simplifying the equation:
P2 = (1.013 x 10^5 Pa) * (0.45 m)^2 / [(2π(0.45 m)^2) + 0.57 m]
Next, we can calculate the change in height (∆h) of the piston:
∆h = h1 - h2
To find the temperature to which the gas should be warmed to raise the piston and dog back to h1, we can use the ideal gas law:
P1V1 / T1 = P2V2 / T2
Since the volume of the gas does not change (∆V = 0), we can simplify the equation to:
P1 / T1 = P2 / T2
Solving for T2:
T2 = T1 * (P2 / P1)
Substituting the given values:
T2 = 294 K * (P2 / 1.013 x 10^5 Pa)
Finally, we can convert the ∆h and T2 to the required units of millimeters and degrees Celsius, respectively.
Note: The calculations involving specific numerical values require additional steps that are omitted in this summary.
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The
momentum of a Boeing 747 jet plane flying at maximum speed is 1.09
x 100 kg•m/s. If the speed was halved, and the mass was tripled,
the new momentum of the plane would be
The speed of the plane is halved and the mass is tripled, the new momentum of the plane would be 163.5 kg·m/s.
The momentum of an object is defined as the product of its mass and velocity. In this case, the momentum of the Boeing 747 jet plane flying at maximum speed is given as 1.09 × 100 kg·m/s.
If the speed of the plane is halved, the new velocity would be half of the original value. Let's call this new velocity v'. The mass of the plane is tripled, so the new mass would be three times the original mass. Let's call this new mass m'.
The momentum of the plane can be calculated using the formula p = mv, where p is the momentum, m is the mass, and v is the velocity.
Since the speed is halved, the new velocity v' is equal to half of the original velocity, so v' = (1/2)v.
Since the mass is tripled, the new mass m' is equal to three times the original mass, so m' = 3m.
The new momentum of the plane, p', can be calculated using the formula p' = m'v':
p' = (3m) × (1/2v) = (3/2)(mv) = (3/2)(1.09 × 100 kg·m/s) = 163.5 kg·m/s.
Therefore, if the speed of the plane is halved and the mass is tripled, the new momentum of the plane would be 163.5 kg·m/s.
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A trrall plaste ball of mass \( m=1.30 \) a ls suspended by a string of length \( 4=17.5 \) \( f=14.5^{\circ} \) argle with the vertical at lnd caber, what is the thet eharge on the bas?"
The trrall plaste ball is suspended by a string of length 4=17.5, forming an angle of 14.5 degrees with the vertical. The task is to determine the charge on the ball.
In the given scenario, the ball is suspended by a string, which means it experiences two forces: tension in the string and the force of gravity. The tension in the string provides the centripetal force necessary to keep the ball in circular motion. The gravitational force acting on the ball can be split into two components: one along the direction of tension and the other perpendicular to it.
By resolving the forces, we find that the component of gravity along the direction of tension is equal to the tension itself. This implies that the magnitude of the tension is equal to the weight of the ball. Using the mass of the ball (m = 1.30), we can calculate its weight using the formula weight = mass × acceleration due to gravity.
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How much would a simple pendulum deflect due to the gravity of a nearby a mountain? As a model of a large mountain, use a sphere of radius R = 2.4 km and mass density = 3000 kg/m3. If a small mass is hung at the end of a string of length 0.80 m at a distance of 3.7 R from the center of the sphere (and assuming the sphere pulls in a horizontal direction on the hanging mass), how far would the small hanging mass deflect under the influence of the sphere's gravitational force? Your answer should be in um (micrometers, 10-6 m):
The deflection of a simple pendulum due to the gravity of a nearby mountain can be determined by calculating the gravitational force exerted by the mountain on the small hanging mass and using it to find the angular displacement of the pendulum.
To begin, let's calculate the gravitational force exerted by the mountain on the small mass. The gravitational force between two objects can be expressed using Newton's law of universal gravitation:
F = G * (m₁ * m₂) / r⁻²
Where F is the gravitational force, G is the gravitational constant (approximately 6.67430 × 10⁻ ¹¹ m³ kg⁻¹ s⁻²), m₁and m ₂ are the masses of the two objects, and r is the distance between their centers.
In this case, the small hanging mass can be considered negligible compared to the mass of the mountain. Thus, we can calculate the force exerted by the mountain on the small mass.
First, let's calculate the mass of the mountain using its volume and density:
V = (4/3) * π * R³
Where V is the volume of the mountain and R is its radius.
Substituting the given values, we have:
V = (4/3) * π * (2.4 km)³
Next, we can calculate the mass of the mountain:
m_mountain = density * V
Substituting the given density of the mountain (3000 kg/m³), we have:
m_mountain = 3000 kg/m³ * V
Now, we can calculate the force exerted by the mountain on the small mass. Since the force is attractive, it will act towards the center of the mountain. Considering that the pendulum's mass is at a distance of 3.7 times the mountain's radius from its center, the force will have a horizontal component.
F_gravity = G * (m_mountain * m_small) / r²
Where F_gravity is the gravitational force, m_small is the mass of the small hanging mass, and r is the distance between their centers.
Substituting the given values, we have:
F_gravity = G * (m_mountain * m_small) / (3.7 * R)²
Next, we need to determine the angular displacement of the pendulum caused by this gravitational force. For small angles of deflection, the angular displacement is directly proportional to the linear displacement.
Using the small angle approximation, we can express the angular displacement (θ) in radians as:
θ = d / L
Where d is the linear displacement of the small mass and L is the length of the pendulum string.
Substituting the given values, we have:
θ = d / 0.80 m
Finally, we can find the linear displacement (d) by multiplying the angular displacement (θ) by the length of the pendulum string (L). Since we want the answer in micrometers (μm), we need to convert the linear displacement from meters to micrometers.
d = θ * L * 10⁶ μm/m
Substituting the given length of the pendulum string (0.80 m) and the calculated angular displacement (θ), we can now solve for the linear displacement (d) in micrometers (μm).
d = θ * 0.80 m * 10⁶ μm/m
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A particle of mass m moves in a three dimensional box with sides L. If the particle is in the first excited level, corresponding to n2 = 6, find
a) energy of particle
b) combination of n1, n2, n3 that would give this energy
c) the wavefunctions for these different states, and
d) the degeneracy of this state
a) The energy of the particle in the first excited level, corresponding to n2 = 6 is 36h² / 8mL².
b) The combination of n1, n2, n3 that would give this energy is (0, 6, 0).
c) The wave function is ψn1, n2, n3 (x,y,z) = √(8/L³)sin((n1πx)/L)sin((n2πy)/L)sin((n3πz)/L).
d) The degeneracy of this state is 1.
a) In quantum mechanics, the energy of a particle in a box is given by E = n²h² / 8mL². In this problem, the particle is in the first excited level corresponding to n2 = 6. We know that n = √6, so the energy of the particle in this state is E = 36h² / 8mL².
b) The particle is excited only in the second direction, so the combination of n1, n2, n3 that would give this energy is (0, 6, 0). c)
The wave function of the particle is given by ψn1, n2, n3 (x,y,z) = √(8/L³)sin((n1πx)/L)sin((n2πy)/L)sin((n3πz)/L).
d) Finally, the degeneracy of this state is 1 since this energy level can only be achieved in one way.
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Near saturation, suppose that the alignment of spins in iron contributes o M = 2.00T to the total magnetic field B. If each electron contributes a magnetic moment of 9.27 × 10−²4 A·m² (one Bohr magneton), about how many electrons per atom contribute to the field? HINT: The total magnetic field is B = Bo + Mo M, where Bo is the externally applied magnetic field and M = xnµp is the magnetic dipoles per volume in the material. Iron contains n = 8.50 × 1028 atoms/m³. x represents the number of electrons per atom that contribute. OA. (a) 1 electron per atom O B. (b) 2 electrons per atom OC. (c) 3 electrons per atom OD. (d) 4 electrons per atom O E. (e) 5 electrons per atom
The magnetic moment is 3 electrons per atom.
Given, M = 2.00T, B = B_o + M_oM
where B_o = externally applied magnetic field , M = xnµp= magnetic dipoles per volume in the material, n = 8.50 × 10^28 atoms/m³.
The magnetic moment of each electron = 9.27 × 10^-24 A·m².
To calculate the number of electrons per atom that contribute to the field, we use the formula:
M = (n × x × µp)Bo + (n × x × µp × M)
The magnetic field is directly proportional to the number of electrons contributing to the field, we can express this relationship as:
n × x = Mo / (µp).
Using the above expression to calculate the value of n × x:n × x = M / (µp) = 2 / (9.27 × 10^-24) = 2.16 × 10^23n = number of atoms/m³.
x = number of electrons/atom
x = (n × x) / n
= 2.16 × 10^23 / 8.5 × 10^28
= 0.2535.
The number of electrons per atom that contribute a magnetic moment of 9.27 × 10−²4 A·m² to the field is approximately 0.25,
Therefore the answer is 0.25 or (c) 3 electrons per atom.
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Which of the following statements for single optic devices are true? Choose all that apply.
All converging optics have a negative focal length.
For virtual images, the object distance is positive and the image distance is positive.
By convention, if the image height is positive then the image is upright.
A magnification of -6 means the image is magnified.
It turns out that virtual images can be created by concave mirrors.
An image with a magnification of 2 is a virtual image.
The correct statements for single optic devices are:
1. For virtual images, the object distance is positive and the image distance is positive.
2. It turns out that virtual images can be created by concave mirrors.
1. For a single optic device, such as a lens or a mirror, the sign convention determines the positive and negative directions. In the sign convention, the object distance (denoted as "do") is positive when the object is on the same side as the incident light, and the image distance (denoted as "di") is positive when the image is formed on the opposite side of the incident light. For virtual images, the object distance is positive and the image distance is positive.
2. Virtual images can indeed be created by concave mirrors. A concave mirror is a converging optic, meaning it can bring parallel incident light rays to a focus. When the object is placed between the focal point and the mirror's surface, a virtual image is formed on the same side as the object. This image is virtual because the reflected rays do not actually converge to form a real image. Instead, they appear to diverge from a virtual point behind the mirror, creating the virtual image.
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A 10 m wide building has a gable shaped roof that is
angled at 23.0° from the horizontal (see the linked
figure).
What is the height difference between the lowest and
highest point of the roof?
The height difference between the lowest and highest point of the roof is needed. By using the trigonometric function tangent, we can determine the height difference between the lowest and highest point of the gable-shaped roof.
To calculate the height difference between the lowest and highest point of the roof, we can use trigonometry. Here's how:
1. Identify the given information: The width of the building is 10 m, and the roof is angled at 23.0° from the horizontal.
2. Draw a diagram: Sketch a triangle representing the gable roof. Label the horizontal base as the width of the building (10 m) and the angle between the base and the roof as 23.0°.
3. Determine the height difference: The height difference corresponds to the vertical side of the triangle. We can calculate it using the trigonometric function tangent (tan).
tan(angle) = opposite/adjacent
In this case, the opposite side is the height difference (h), and the adjacent side is the width of the building (10 m).
tan(23.0°) = h/10
Rearrange the equation to solve for h:
h = 10 * tan(23.0°)
Use a calculator to find the value of tan(23.0°) and calculate the height difference.
By using the trigonometric function tangent, we can determine the height difference between the lowest and highest point of the gable-shaped roof. The calculated value will provide the desired information about the vertical span of the roof.
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1. 7points Can two displacement vectors of the same length have a vector sum of zero? Explain-Draw a graph
No, two displacement vectors of the same length cannot have a vector sum of zero.
If two vectors have the same length but their directions are not opposite, their vector sum will always result in a non-zero vector. When we add vectors graphically, we can represent each vector as an arrow and place them tip-to-tail. If the resulting vector ends at the origin (zero), it means the vector sum is zero. However, since the two vectors have the same length, their arrows will always be parallel, and placing them tip-to-tail will result in a longer vector pointing in a specific direction. Thus, the vector sum can never be zero for two non-opposite vectors of the same length.
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No, two displacement vectors of the same length cannot have a vector sum of zero.
If two vectors have the same length but their directions are not opposite, their vector sum will always result in a non-zero vector.
When we add vectors graphically, we can represent each vector as an arrow and place them tip-to-tail. If the resulting vector ends at the origin (zero), it means the vector sum is zero.
However, since the two vectors have the same length, their arrows will always be parallel, and placing them tip-to-tail will result in a longer vector pointing in a specific direction. Thus, the vector sum can never be zero for two non-opposite vectors of the same length.
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Identify three things in Figure 5 that help make the skier complete the race faster. Figure 5
This enables the skier to make quick and accurate turns, which is especially important when skiing downhill at high speeds.
In Figure 5, the following are the three things that help the skier complete the race faster:
Reduced air resistance: The skier reduces air resistance by crouching low, which decreases air drag. This enables the skier to ski faster and more aerodynamically. This is demonstrated by the skier in Figure 5 who is crouching low to reduce air resistance.
Rounded ski tips: Rounded ski tips help the skier to make turns more quickly. This is because rounded ski tips make it easier for the skier to glide through the snow while turning, which reduces the amount of time it takes for the skier to complete a turn.
Sharp edges: Sharp edges on the skier’s skis allow for more precise turning and edge control.
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An ice dancer with her arms stretched out starts into a spin with an angular velocity of 2.2 rad/s. Her moment of inertia with her arms stretched out is 2.74kg m? What is the difference in her rotational kinetic energy when she pulls in her arms to make her moment of inertia 1.54 kg m2?
The difference in rotational kinetic energy when the ice dancer pulls in her arms from a moment of inertia of 2.74 kg m² to 1.54 kg m² is 0.998 Joules.
When the ice dancer pulls in her arms, her moment of inertia decreases, resulting in a change in rotational kinetic energy. The formula for the difference in rotational kinetic energy (ΔK) is given by ΔK = ½ * (I₂ - I₁) * (ω₂² - ω₁²), where I₁ and I₂ are the initial and final moments of inertia, and ω₁ and ω₂ are the initial and final angular velocities.
Given I₁ = 2.74 kg m², I₂ = 1.54 kg m², and ω₁ = 2.2 rad/s, we can calculate ω₂ using the conservation of angular momentum, I₁ * ω₁ = I₂ * ω₂. Solving for ω₂ gives ω₂ = (I₁ * ω₁) / I₂.
Substituting the values into the formula for ΔK, we have ΔK = ½ * (I₂ - I₁) * [(I₁ * ω₁ / I₂)² - ω₁²].
Performing the calculations, we find ΔK ≈ 0.998 Joules. This means that when the ice dancer pulls in her arms, the rotational kinetic energy decreases by approximately 0.998 Joules.
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Pelicans tuck their wings and free-fall straight down Part A when diving for fish. Suppose a pelican starts its dive from a height of 20.0 m and cannot change its If it takes a fish 0.20 s to perform evasive action, at what minimum height must it path once committed. spot the pelican to escape? Assume the fish is at the surface of the water. Express your answer using two significant figures.
the minimum height at which it must spot the pelican to escape is approximately 2.02 s * 0.20 s = 0.404 m, which can be rounded to 0.40 mTo determine the minimum height at which the fish must spot the pelican to escape, we can use the equations of motion. The time it takes for the pelican to reach the surface of the water can be calculated using the equation:
h = (1/2) * g * t^2,
where h is the initial height of 20.0 m, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken by the pelican to reach the surface.
Rearranging the equation to solve for t, we have:
t = sqrt(2h / g).
Substituting the given values into the equation, we get:
t = sqrt(2 * 20.0 m / 9.8 m/s^2) ≈ 2.02 s.
Since the fish has only 0.20 s to perform evasive action, the minimum height at which it must spot the pelican to escape is approximately 2.02 s * 0.20 s = 0.404 m, which can be rounded to 0.40 m (two significant figures).
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A grindstone is accelerated from rest to 32 rad/s in 0.40 s. (a) What is the angular acceleration in rad/s^2? (b) How many revolutions does it go through in the process?
The angular acceleration is 80 rad/s^2, and the grindstone goes through approximately 1.02 revolutions during the acceleration process.
To determine the angular acceleration and the number of revolutions, we are given the initial angular velocity, final angular velocity, and the time taken for acceleration.
The explanation of the answers will be provided in the second paragraph.
(a) The angular acceleration (α) can be calculated using the formula:
α = (ωf - ωi) / t
where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time taken for acceleration.
Plugging in the given values, we have:
α = (32 rad/s - 0 rad/s) / 0.40 s
α = 80 rad/s^2
(b) To determine the number of revolutions, we can use the formula:
θ = ωi * t + (1/2) * α * t^2
where θ is the angular displacement in radians, ωi is the initial angular velocity, t is the time taken for acceleration, and α is the angular acceleration.
Plugging in the given values, we have:
θ = 0 rad/s * 0.40 s + (1/2) * 80 rad/s^2 * (0.40 s)^2
θ = 6.4 rad
To convert radians to revolutions, we divide by 2π:
θ (in revolutions) = 6.4 rad / (2π rad/rev)
θ (in revolutions) ≈ 1.02 rev
In summary, the angular acceleration is 80 rad/s^2, and the grindstone goes through approximately 1.02 revolutions during the acceleration process.
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Greta took an IQ test and scored high in knowledge and vocabulary. Which of the following statements BEST describes Greta’s results?
Answer:
Greta scored high in knowledge and vocabulary on the IQ test.
Explanation:
This statement highlights Greta's strengths in knowledge and vocabulary specifically, indicating that she performed well in these areas during the test. However, it does not provide information about her overall IQ score or her performance in other cognitive domains that may have been assessed in th
With her advanced education Jacky decides to become a nuclear engineer for the Navy and is currently on a submarine off the coast of North Korea. If the pressure of the water outside of Jacky's submarine is 32 atm, how deep is her submarine?
[the density of sea water is 1,025 kg/m^3]
Group of answer choices
A. 311.7 m
B. 51.1 m
C. 117.6 m
D. 277.2 m
Jacky is a nuclear engineer who is currently on a submarine off the coast of North Korea. If the pressure of the water outside of Jacky's submarine is 32 atm, how deep is her submarine the density of sea water is 1,025 kg/m³.
The pressure of a liquid is directly proportional to its depth in the liquid. Furthermore, the higher the density of the fluid, the higher the pressure exerted. We'll use the following formula :P = ρgh Where:P = pressure in pascalsρ = density of the fluid in kg/m³g = acceleration due to gravity, which is 9.8 m/s²h = height of the fluid column in meters
The pressure at any depth h below the surface is given by the formula:
P = Patm + ρghWhere:Patm = atmospheric pressureρ = density of the fluidg = acceleration due to gravity,
which is 9.8 m/s²h = depth of the liquid column The pressure outside the submarine is given as 32 atm. This is equivalent to
:P = 32 atm × 1.013 × 10⁵ Pa/atm = 3.232 × 10⁶ PaWe will use the formula ,P = Patm + ρgh
to determine the depth of the submarine.
Patm = atmospheric pressure =
1 atm = 1.013 × 10⁵ Paρ = density of the sea water = 1025 kg/m³g =
acceleration due to gravity = 9.8 m/s²h = depth of the submarine
By substituting the values,
we get3.232 × 10⁶ Pa = 1.013 × 10⁵ Pa + (1025 kg/m³ × 9.8 m/s² × h)Solving for h we get h = 277.23
the depth of the submarine is 277.23 m Option D is the correct answer.
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an object moves up and down in simple harmonic motion with an amplitude of 4.46 cm and a frequency of 1.65 Hz. what is the max speed of the object ?
The maximum speed of an object that moves up and down in simple harmonic motion with an amplitude of 4.46 cm and a frequency of 1.65 Hz is 0.293 m/s.
Simple harmonic motion is defined as the motion of an object back and forth around its mean position. For example, when a pendulum swings, it exhibits simple harmonic motion because it moves back and forth around its equilibrium position.
The maximum speed of an object undergoing simple harmonic motion is given by the formula:
vmax = Aω
where A is the amplitude of the motion and ω is the angular frequency.ω can be determined using the formula
ω = 2πf
where f is the frequency of the motion.
Using these formulas, we can determine the maximum speed of the object:
vmax = Aω
vmax = 0.0446 m x (2π x 1.65 Hz)
vmax ≈ 0.293 m/s
Therefore, the maximum speed of the object is 0.293 m/s.
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Q4: Let's combine our observations on the gravitational force, velocity and path and provide a full explanation on why the velocity and the path of the Earth around the Sun change drastically when we double the mass of the Sun but not when we double the mass of the Earth.
When we double the mass of the Sun, the increased gravitational force leads to a decrease in the Earth's acceleration, resulting in a slower velocity and a larger orbit. On the other hand, when we double the mass of the Earth, the gravitational force does not change significantly,
When considering the gravitational force, velocity, and path of the Earth around the Sun, we need to take into account the fundamental principles of gravitational interactions described by Newton's law of universal gravitation and the laws of motion.
Newton's Law of Universal Gravitation:
According to Newton's law of universal gravitation, the force of gravitational attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers of mass.
F = G × (m1 × m2) / r²
Where:
F is the gravitational force between the two objects,
G is the gravitational constant,
m1 and m2 are the masses of the two objects, and
r is the distance between their centers of mass.
Laws of Motion:
The motion of an object is determined by Newton's laws of motion, which include the concepts of inertia, force, and acceleration.
Newton's First Law (Law of Inertia): An object at rest or in uniform motion will remain in that state unless acted upon by an external force.
Newton's Second Law: The force acting on an object is equal to the mass of the object multiplied by its acceleration.
Newton's Third Law: For every action, there is an equal and opposite reaction.
When we double the mass of the Sun:
By doubling the mass of the Sun, the gravitational force between the Earth and the Sun increases due to the direct proportionality between the force and the masses. The increased gravitational force leads to a higher acceleration experienced by the Earth.
According to Newton's second law (F = m ×a), for a given force, an object with a larger mass will experience a smaller acceleration. Therefore, with the doubled mass of the Sun, the Earth's acceleration decreases compared to the original scenario.
As a result, the Earth's velocity and path around the Sun will change drastically. The decreased acceleration causes the Earth to move at a slower velocity, resulting in a longer orbital period and a larger orbital radius. The Earth will take more time to complete one revolution around the Sun, and its path will be wider due to the decreased curvature of the orbit.
When we double the mass of the Earth:
When we double the mass of the Earth, the gravitational force between the Earth and the Sun does not change significantly. Although the gravitational force is affected by the mass of both objects, doubling the Earth's mass while keeping the Sun's mass constant does not lead to a substantial change in the gravitational force.
According to Newton's second law, the acceleration of an object is directly proportional to the applied force and inversely proportional to the mass. Since the gravitational force remains relatively constant, doubling the mass of the Earth leads to a decrease in the Earth's acceleration.
Consequently, the Earth's velocity and path around the Sun are not drastically affected by doubling its mass. The change in acceleration is relatively small, resulting in a slightly slower velocity and a slightly wider orbit, but these changes are not significant enough to cause a drastic alteration in the Earth's orbital dynamics.
In summary, when we double the mass of the Sun, the increased gravitational force leads to a decrease in the Earth's acceleration, resulting in a slower velocity and a larger orbit. On the other hand, when we double the mass of the Earth, the gravitational force does not change significantly, and the resulting small decrease in acceleration only causes a minor variation in the Earth's velocity and path.
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24. (True/False) The tangential acceleration for a point on a solid rotating object depends on the point's radial distance from the axis of rotation. 25. (True/False) Kepler's third law relates the square of a planet's orbital period to the square of its orbital distance from the Sun. 26. (True/False) Increasing the distance between the rotation axis and the point at which a force is applied will increase the torque (assuming the angle of application is kept fixed). 27. (True/False) The moment of inertia for an object is independent of the location of the rotation axis. 28. (True/False) The continuity equation for fluid flow through a pipe relates the cross-sectional areas and speeds at two different points in the pipe. 29. (True/False) Heat flows between two objects at the same temperature in thermal contact if one object is larger than the other. 30. (True/False) A material's specific heat quantifies the energy per unit mass needed to induce a phase change. 31. The first law of thermodynamics relates the total change in a system's internal energy to energy transfers due to heat and work.
24. False. The tangential acceleration for a point on a solid rotating object does not depend on the point's radial distance from the axis of rotation. It is the same for all points located at the same radius.
25. True. Kepler's third law relates the square of a planet's orbital period to the square of its orbital distance from the Sun. It is also called the law of periods.
26. True. Increasing the distance between the rotation axis and the point at which a force is applied will increase the torque, assuming the angle of application is kept fixed. Torque is equal to the product of force and the perpendicular distance of the line of action of force from the axis of rotation.
27. True. The moment of inertia for an object is independent of the location of the rotation axis. It is the same no matter where the axis is located in the object.
28. True. The continuity equation for fluid flow through a pipe relates the cross-sectional areas and speeds at two different points in the pipe. The product of cross-sectional area and speed is constant throughout the pipe.
29. False. Heat does not flow between two objects at the same temperature in thermal contact, regardless of the size of the objects. Heat flows from a higher temperature to a lower temperature.
30. False. A material's specific heat quantifies the energy required to change the temperature of the unit mass of the material, not to induce a phase change.
31. True. The first law of thermodynamics relates the total change in a system's internal energy to energy transfers due to heat and work. It is also known as the law of conservation of energy.
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ght of wavelength 590.0 nm illuminates a slit of width 0.74 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum? 2 m (b) Calculate the width of the central maximum. 20 How is the width of the central maximum related to the distance from the central maximum to the first minimum? find the width of the central maximum. mm
To find the distance from the slit to the screen, we can use the formula for the location of the first minimum in the diffraction pattern: y = (λ * L) / d
y is the distance from the central maximum to the first minimum, λ is the wavelength of the light (590.0 nm = 5.9 * 10^-7 m), L is the distance from the slit to the screen (which we need to find), and d is the width of the slit (0.74 mm = 7.4 * 10^-4 m). Plugging in the values, we have:
0.93 * 10^-3 m = (5.9 * 10^-7 m) * L / (7.4 * 10^-4 m)
Solving for L, we get:
L = (0.93 * 10^-3 m) * (7.4 * 10^-4 m) / (5.9 * 10^-7 m) ≈ 1.17 m
So, the distance from the slit to the screen should be approximately 1.17 m.
(b) The width of the central maximum can be calculated using the formula:
w = (λ * L) / d
Where:
w is the width of the central maximum.
Plugging in the values, we have:
w = (5.9 * 10^-7 m) * (1.17 m) / (7.4 * 10^-4 m) ≈ 9.3 * 10^-4 m
So, the width of the central maximum is approximately 9.3 * 10^-4 m or 0.93 mm.
The width of the central maximum is related to the distance from the central maximum to the first minimum by the formula w = 2 * y, where y is the distance from the central maximum to the first minimum. Therefore, the width of the central maximum is twice the distance from the central maximum to the first minimum.
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A worker lifts a box upward from the floor and then carries it across the warehouse. When is he doing work? while lifting the box from the floor while carrying the box across the warehouse while standing in place with the box at no time during the process A baseball player drops the ball from his glove. At what moment is the ball's kinetic energy the greatest? when the baseball player is holding the ball at the ball's highest point before beginning to fall just before the ball hits the ground the moment the ball leaves the baseball player's glove
A worker lifts a box upwards from the floor and then carries it across the warehouse. At the moment the ball leaves the baseball player's glove, the kinetic energy of the ball is the greatest.
The worker is doing work while lifting the box from the floor and carrying the box across the warehouse. A worker lifts a box upward from the floor and then carries it across the warehouse. When he is lifting the box from the floor and carrying the box across the warehouse, he is doing work. According to physics, work done when force is applied to an object to move it over a distance in the same direction as the applied force.
while lifting the box from the floor and while carrying the box across the warehouse, the worker is doing work. Thus, the worker is doing work while he is lifting the box from the floor and carrying the box across the warehouse. The kinetic energy of the ball is the greatest at the moment the ball leaves the baseball player's glove. A baseball player drops the ball from his glove. At the moment the ball leaves the baseball player's glove, the kinetic energy of the ball is the greatest.
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In your own words, explain the difference between a wave and a vibration.
Vibrations are localized oscillations, while waves are disturbances that propagate through a medium or space.
1. Vibration:
A vibration refers to a repetitive back-and-forth or oscillating motion of an object or a system around a fixed position.
It involves the periodic movement of particles or components within an object or medium.
The motion of the object or system can be linear or rotational.
Key characteristics of vibrations include:
- Periodicity: Vibrations occur with a regular pattern or cycle.
- Amplitude: It represents the maximum displacement or distance from the equilibrium position that an object or particle achieves during vibration.
- Frequency: It is the number of complete cycles or oscillations per unit of time, typically measured in hertz (Hz).
- Energy transfer: Vibrations often involve the transfer of energy from one object or medium to another.
Examples of vibrations include the oscillation of a pendulum, the back-and-forth motion of a guitar string, or the movement of atoms in a solid material when subjected to thermal energy.
2. Wave:
A wave refers to the propagation of energy through a medium or space without a net displacement of the medium itself.
Waves transmit energy by causing a disturbance or oscillation to propagate through particles or fields.
Key characteristics of waves include:
- Propagation: Waves travel through space or a medium, transferring energy from one location to another.
- Disturbance: Waves are created by a disturbance or oscillation that sets particles or fields in motion.
- Wavelength: It is the distance between two corresponding points on a wave, such as the distance between two peaks or two troughs.
- Amplitude: It represents the maximum displacement of particles or the maximum value of the wave's quantity (e.g., amplitude of displacement in a water wave or amplitude of oscillation in a sound wave).
- Frequency: It is the number of complete cycles or oscillations of a wave that occur per unit of time, measured in hertz (Hz).
Examples of waves include electromagnetic waves (such as light waves and radio waves), sound waves, water waves, seismic waves, and more.
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Consider the following statements: T/F?
The number 9800. has two significant figures. The number 9.8x10^9 has two significant figures. The number 9.80x10^9 has two significant figures. The number 9800 can have 2, 3, or 4 significant figures, depending on the significance of the zeros. The number 9800. has four significant figures. True The number 9.800x10^9 has four significant figures
1. The number 9800. has two significant figures. False
The number 9800. has four significant figures. As there is a decimal point after 9800, this indicates that the trailing zero (the zero after 9800) is significant.
2. The number 9.8x10^9 has two significant figures. False
The number 9.8x10^9 has two significant figures in the coefficient. The exponent (10^9) is not significant.
3. The number 9.80x10^9 has two significant figures. False
The number 9.80x10^9 has three significant figures in the coefficient. The exponent (10^9) is not significant.
4. The number 9800 can have 2, 3, or 4 significant figures, depending on the significance of the zeros. True
For example, if 9800 is measured, it has two significant figures. If it is written to two decimal places (9800.00), it has six significant figures.
5. The number 9.800x10^9 has four significant figures. True
The number 9.800x10^9 has four significant figures in the coefficient. The exponent (10^9) is not significant.
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A ball falls from height of 19.0 m, hits the floor, and rebounds vertically upward to height of 15.0 m. Assume that Mball = 0.290 kg.
What is the impulse (in kg • m/s) delivered to the ball by the floor?
The impulse is approximately -9.94432 kg * m/s.
To find the impulse delivered to the ball by the floor, we can use the principle of conservation of momentum.
The impulse is equal to the change in momentum of the ball.
The change in momentum of the ball can be calculated as the final momentum minus the initial momentum.
Momentum (p) is given by the product of mass (m) and velocity (v):
p = m * v
Let's assume that the initial velocity of the ball is u and the final velocity after rebounding is v.
Initial momentum = m * u
Final momentum = m * v
Since the ball falls vertically downward, the initial velocity (u) is positive and the final velocity (v) after rebounding is upward, so it is negative.
The change in momentum is:
Change in momentum = Final momentum - Initial momentum = m * v - m * u
Now, let's calculate the velocities:
The velocity just before hitting the floor can be found using the equation of motion for free fall:
v^2 = u^2 + 2 * a * s
Here, u is the initial velocity (which is 0 since the ball is initially at rest), a is the acceleration due to gravity (approximately 9.8 m/s^2), and s is the distance fallen (19.0 m).
v^2 = 0 + 2 * 9.8 * 19.0
v^2 = 372.4
v ≈ √372.4
v ≈ 19.28 m/s
The velocity after rebounding is given as -15.0 m/s (since it is upward).
Now we can calculate the change in momentum:
Change in momentum = m * v - m * u
Change in momentum = 0.290 kg * (-15.0 m/s) - 0.290 kg * (19.28 m/s)
Change in momentum ≈ -4.35 kg * m/s - 5.59432 kg * m/s
Change in momentum ≈ -9.94432 kg * m/s
The impulse delivered to the ball by the floor is equal to the change in momentum, so the impulse is approximately -9.94432 kg * m/s.
The negative sign indicates that the direction of the impulse is opposite to the initial momentum of the ball, as the ball rebounds upward.
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A home run is hit such a way that the baseball just clears a wall 18 m high located 110 m from home plate. The ball is hit at an angle of 38° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1 m above the ground. The acceleration of gravity is 9.8 m/s2. What is the initial speed of the ball? Answer in units of m/s. Answer in units of m/s
The initial speed of the ball is approximately 35.78 m/s.
To find the initial speed of the ball, we can analyze the vertical and horizontal components of its motion separately.
Height of the wall (h) = 18 m
Distance from home plate to the wall (d) = 110 m
Launch angle (θ) = 38°
Initial height (h0) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Analyzing the vertical motion:
The ball's vertical motion follows a projectile trajectory, starting at an initial height of 1 m and reaching a maximum height of 18 m.
The equation for the vertical displacement (Δy) of a projectile launched at an angle θ is by:
Δy = h - h0 = (v₀ * sinθ * t) - (0.5 * g * t²)
At the highest point of the trajectory, the vertical velocity (v_y) is zero. Therefore, we can find the time (t) it takes to reach the maximum height using the equation:
v_y = v₀ * sinθ - g * t = 0
Solving for t:
t = (v₀ * sinθ) / g
Substituting this value of t back into the equation for Δy, we have:
h - h0 = (v₀ * sinθ * [(v₀ * sinθ) / g]) - (0.5 * g * [(v₀ * sinθ) / g]²)
Simplifying the equation:
17 = (v₀² * sin²θ) / (2 * g)
Analyzing the horizontal motion:
The horizontal distance traveled by the ball is equal to the distance from home plate to the wall, which is 110 m.
The horizontal displacement (Δx) of a projectile launched at an angle θ is by:
Δx = v₀ * cosθ * t
Since we have already solved for t, we can substitute this value into the equation:
110 = (v₀ * cosθ) * [(v₀ * sinθ) / g]
Simplifying the equation:
110 = (v₀² * sinθ * cosθ) / g
Finding the initial speed (v₀):
We can now solve the two equations obtained from vertical and horizontal motion simultaneously to find the value of v₀.
From the equation for vertical displacement, we have:
17 = (v₀² * sin²θ) / (2 * g) ... (equation 1)
From the equation for horizontal displacement, we have:
110 = (v₀² * sinθ * cosθ) / g ... (equation 2)
Dividing equation 2 by equation 1:
(110 / 17) = [(v₀² * sinθ * cosθ) / g] / [(v₀² * sin²θ) / (2 * g)]
Simplifying the equation:
(110 / 17) = 2 * cosθ / sinθ
Using the trigonometric identity cosθ / sinθ = cotθ, we have:
(110 / 17) = 2 * cotθ
Solving for cotθ:
cotθ = (110 / 17) / 2 = 6.470588
Taking the inverse cotangent of both sides:
θ = arccot(6.470588)
Using a calculator, we find:
θ ≈ 9.24°
Finally, we can substitute the value of θ into either equation 1 or equation 2 to solve for v₀. Let's use equation 1:
17 = (v₀² * sin²(9.24°)) /
Rearranging the equation and solving for v₀:
v₀² = (17 * 2 * 9.8) / sin²(9.24°)
v₀ = √[(17 * 2 * 9.8) / sin²(9.24°)]
Calculating this expression using a calculator, we find:
v₀ ≈ 35.78 m/s
Therefore, the initial speed of the ball is approximately 35.78 m/s.
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Two beakers of water are on the lab table. One beaker has 30 g of water at 80∘
C and the other has 80 g at 30 ∘C. Which one would require more thermal energy to raise its temperature from 0∘C to its present temperature? Neither would require thermal energy to increase its temperature. Both would require the same amount of thermal energy. We can't tell until we know the specific heat. The 30 g beaker. The 80 g beaker.
The answer to the given problem is the beaker that has 30g of water at 80 °C. This requires more thermal energy to raise its temperature from 0 °C to its present temperature.
Let's recall the formula to calculate the amount of thermal energy required to raise the temperature of a substance.Q = m × c × ΔT where,Q = the amount of heatm = mass of the substancec = specific heat of the substance. ΔT = change in temperature. From the given problem, we have two beakers of water with different masses and temperatures. Therefore, the amount of thermal energy required to raise their temperatures from 0 °C to their current temperature is different. We have;Q1 = m1 × c × ΔT1Q2 = m2 × c × ΔT2 where,m1 = 30g and ΔT1 = 80 - 0 = 80 °Cm2 = 80g and ΔT2 = 30 - 0 = 30 °C. Now we compare Q1 and Q2 to determine which beaker would require more thermal energy. Q1 = m1 × c × ΔT1 = 30g × c × 80 °CQ2 = m2 × c × ΔT2 = 80g × c × 30 °C. Comparing Q1 and Q2, we have;Q1 > Q2. Therefore, the beaker that has 30g of water at 80 °C requires more thermal energy to raise its temperature from 0 °C to its present temperature than the beaker with 80g at 30 °C.
Thus , the answer is the 30g beaker requires more thermal energy to raise its temperature from 0 °C to its present temperature than the 80g beaker.
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Real images formed by a spherical mirror are always: A. on the side of the mirror opposite the source B. on the same side of the mirror as the source but closer to the mirror than the source C. on the same side of the mirror as the source but never any further from the mirror than the focal point D. on the same side of the mirror as the source but never any closer to the mirror than the focal point E. none of the above
The correct option is D. on the same side of the mirror as the source but never any closer to the mirror than the focal point.
A spherical mirror is a mirror that has a spherical shape like a ball. A spherical mirror is either concave or convex. The mirror has a center of curvature (C), a radius of curvature (R), and a focal point (F).
When a ray of light traveling parallel to the principal axis hits a concave mirror, it is reflected through the focal point. It forms an image that is real, inverted, and magnified when the object is placed farther than the focal point. If the object is placed at the focal point, the image will be infinite.
When the object is placed between the focal point and the center of curvature, the image will be real, inverted, and magnified, while when the object is placed beyond the center of curvature, the image will be real, inverted, and diminished.
In the case of a convex mirror, when a ray of light parallel to the principal axis hits the mirror, it is reflected as if it came from the focal point. The image that is formed by a convex mirror is virtual, upright, and smaller than the object.
The image is always behind the mirror, and the image distance (di) is negative. Therefore, the correct option is D. on the same side of the mirror as the source but never any closer to the mirror than the focal point.
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Type your answers in all of the blanks and submit S ⋆⋆ A cylindrical glass beaker has an inside diameter of 8.0 cm and a mass of 200 g. It is filled with water to a height of 5.0 cm. The water-filled beaker is placed on a weight scale. A solid cylinder of aluminum that is 8.0 cm tall and has a radius of 2.0 cm is tied to a string. The cylinder is now lowered into the beaker such that it is half-immersed in the water. Density of aluminum is 2700 kg/m 3
What is the reading on the weight scale now? N What is the tension in the string? N
The reading on the weight scale now is 4.295 N and the tension in the string is 0.189 N.
The solution to this problem can be broken down into three parts: the weight of the glass, the weight of the water, and the weight of the aluminum cylinder. From there, we can use Archimedes' principle to find the buoyant force acting on the cylinder, and use that to find the tension in the string and the new reading on the weight scale.
Let's begin.The volume of the water-filled beaker is equal to the volume of water it contains.
Therefore, we can calculate the volume of water as follows:
V = πr²h
πr²h = π(0.04 m)²(0.05 m),
π(0.04 m)²(0.05 m) = 2.0 x 10⁻⁵ m³.
We can also calculate the mass of the water as follows:
m = ρV ,
ρV = (1000 kg/m³)(2.0 x 10⁻⁵ m³) ,
(1000 kg/m³)(2.0 x 10⁻⁵ m³) = 0.02 kg.
Next, we can find the weight of the glass using its mass and the acceleration due to gravity:
w = mg,
mg = (0.2 kg)(9.81 m/s²) ,
(0.2 kg)(9.81 m/s²) = 1.962 N.
To find the weight of the aluminum cylinder, we first need to calculate its volume:
V = πr²h
= π(0.02 m)²(0.08 m) ,
π(0.02 m)²(0.08 m) = 1.005 x 10⁻⁴ m³.
We can then find its mass using its volume and density:
m = ρV,
ρV = (2700 kg/m³)(1.005 x 10⁻⁴ m³),
(2700 kg/m³)(1.005 x 10⁻⁴ m³) = 0.027135 kg.
Finally, we can find the weight of the aluminum cylinder:
w = mg ,
mg = (0.027135 kg)(9.81 m/s²),
(0.027135 kg)(9.81 m/s²) = 0.266 N.
Now that we have found the weights of the glass, water, and aluminum cylinder, we can add them together to find the total weight of the system:
1.962 N + 0.02 kg(9.81 m/s²) + 0.266 N = 4.295 N.
This is the new reading on the weight scale. However, we still need to find the tension in the string.To do this, we need to find the buoyant force acting on the aluminum cylinder. The volume of water displaced by the cylinder is equal to the volume of the cylinder that is submerged in the water. This volume can be found by multiplying the cross-sectional area of the cylinder by the height of the water level:
Vd = Ah ,
Ah = πr²h/2 ,
πr²h/2 = π(0.02 m)²(0.025 m) ,
π(0.02 m)²(0.025 m) = 7.854 x 10⁻⁶ m³.
Since the density of water is 1000 kg/m³, we can find the buoyant force using the following formula:
Fb = ρgVd,
ρgVd = (1000 kg/m³)(9.81 m/s²)(7.854 x 10⁻⁶ m³),
(1000 kg/m³)(9.81 m/s²)(7.854 x 10⁻⁶ m³) = 0.077 N.
The tension in the string is equal to the weight of the aluminum cylinder minus the buoyant force acting on it:
T = w - Fb,
w - Fb = 0.266 N - 0.077 N,
0.266 N - 0.077 N = 0.189 N.
Therefore, the reading on the weight scale now is 4.295 N and the tension in the string is 0.189 N.
The reading on the weight scale now is 4.295 N and the tension in the string is 0.189 N.
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