Capacitive reactance (Xc) is a measure of the opposition to the flow of alternating current (AC) through a capacitor. Both capacitors have a capacitive reactance of 40 Ω, and the AC voltage source has a frequency of almost 80 Hz.
Capacitive reactance arises due to the behavior of a capacitor in an AC circuit. A capacitor stores electrical energy in an electric field between its plates when it is charged. When an AC voltage is applied to a capacitor, the voltage across the capacitor changes with the frequency of the AC signal. As the frequency increases, the capacitor has less time to charge and discharge, resulting in a higher opposition to the flow of current.
To solve this problem, we can use the formula for capacitive reactance (Xc) in an AC circuit:
[tex]Xc = 1 / (2\pi fC)[/tex]
Where:
Xc is the capacitive reactance in ohms (Ω),
π is a mathematical constant (approximately 3.14159),
f is the frequency of the AC voltage source in hertz (Hz),
C is the capacitance in farads (F).
Let's solve for the frequency of the AC voltage source and the capacitive reactance for each capacitor:
For the 100-pF capacitor:
Given:
[tex]C = 100 pF = 100 * 10^{-12} F\\X_c = 20 \Omega[/tex]
[tex]20 \Omega = 1 / (2\pi f * 100 * 10^{-12} F)[/tex]
Solving for f:
[tex]f = 1 / (2\pi * 20 \Omega * 100 * 10^{-12} F)\\f = 79577.68 Hz = 80 kHz[/tex]
Therefore, the frequency of the AC voltage source is approximately 80 kHz for the 100-pF capacitor.
For the 50-μF capacitor:
[tex]C = 50 \mu F = 50 * 10^{-6} F[/tex]
We want to find the capacitive reactance (Xc) for this capacitor:
[tex]X_c = 1 / (2\pi f * 50 * 10^{-6} F)[/tex]
To show that the capacitive reactance will be 40 Ω, we substitute the value of Xc into the equation:
[tex]40 \Omega = 1 / (2\pi f * 50 * 10^{-6}F)\\f = 1 / (2\pi * 40 \Omega * 50 * 10^{-6} F)\\f = 79577.68 Hz = 80 kHz[/tex]
Again, the frequency of the AC voltage source is approximately 80 kHz for the 50-μF capacitor.
Hence, both capacitors have a capacitive reactance of 40 Ω, and the AC voltage source has a frequency of almost 80 Hz.
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The 50-µF capacitor has a capacitive reactance twice as that of the 100-pF capacitor.
Given information, The capacitive reactance of a 100-pF capacitor is 20 Ω
The capacitive reactance of a 50-µF capacitor is to be determined
The frequency of the AC voltage source is almost 80 Hz
The capacitive reactance of a capacitor is given by the relation, XC = 1 / (2πfC)
WhereXC = Capacitive reactance, C = Capacitance, f = Frequency
On substituting the given values for the 100-pF capacitor, the frequency of the AC voltage source is found to be,20 = 1 / (2πf × 100 × 10⁻¹²)⇒ f = 1 / (2π × 20 × 100 × 10⁻¹²) = 7.957 Hz
On substituting the given values for the 50-µF capacitor, its capacitive reactance is found to be, XC = 1 / (2πfC)⇒ XC = 1 / (2π × 7.957 × 50 × 10⁻⁶) = 39.88 Ω ≈ 40 Ω
The capacitive reactance of the 50-µF capacitor is 40 Ω and the frequency of the AC voltage source is almost 80 Hz, which was calculated to be 7.957 Hz for the 100-pF capacitor.
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A charge of +12 nC is placed on the x-axis at x = 4.4 m, and a charge of -25 nC is placed at x = -5.6 m. What is the magnitude of the electric field at the origin? Give your answer to a decimal place.
The magnitude of the electric field at the origin, rounded to one decimal place, is approximately 7.2 × 10⁶ N/C.
We need to calculate the electric field contribution from each charge and then add them together to find the net electric field at the origin.
Given:
Charge Q1 = +12 nC
Position x1 = 4.4 m
Charge Q2 = -25 nC
Position x2 = -5.6 m
Electrostatic constant k ≈ 8.99 × 10⁹ Nm²/C²
First, let's calculate the electric field contribution from Q1:
E1 = (8.99 × 10⁹ Nm²/C²) * (12 × 10⁻⁹ C) / (4.4 m)²
Substituting the values and performing the calculation:
E1 = 2.40 × 10⁶ N/C
Next, we calculate the electric field contribution from Q2:
E2 = (8.99 × 10⁹ Nm²/C²) * (-25 × 10⁻⁹) C) / (5.6 m)²
Substituting the values and performing the calculation:
E2 = -9.59 × 10⁶ N/C
Now, let's find the net electric field at the origin by summing the contributions:
E_net = E1 + E2
Substituting the values and performing the calculation:
E_net = (2.40 × 10⁶ N/C) + (-9.59 × 10⁶ N/C)
E_net = -7.19 × 10⁶ N/C
Finally, we take the magnitude of E_net to find the absolute value of the electric field at the origin:
|E_net| = |-7.19 × 10⁶ N/C|
|E_net| = 7.19 × 10⁶ N/C
After calculating the net electric field at the origin as -7.19 × 10⁶ N/C, rounding to one decimal place gives us:
|E_net| ≈ 7.2 × 10⁶ N/C
Therefore, the magnitude of the electric field at the origin, rounded to one decimal place, is approximately 7.2 × 10⁶ N/C.
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Problem 8.44 A centrifuge rotor rotating at 9800 rpm shut off and is eventually brought uniformly to rest by a frictional torque of 1.91 m N. Part A If the mass of the rotor is 4.16 kg and it can be approximated as a solid cylinder of radius 0.0440 m, through how many revolutions will the rotor turn before coming to rest? Express your answer to three significant figures. VE ΑΣΦ N = 71.6 Submit Part B ! You have already submitted this answer. Enter a new answer. No credit lost. Try again. D Previous Answers Request Answer How long will it take? Express your answer to three significant figures and include the appropriate units. t = 0.885 μÅ Provide Feedback S Submit Previous Answers Request Answer ? ? X Incorrect; Try Again; 5 attempts remaining revolutions
The centrifuge rotor, with a mass of 4.16 kg and a radius of 0.0440 m, comes to rest after a frictional torque of 1.91 mN is applied.
To find the number of revolutions the rotor will turn before coming to rest, we can use the relationship between torque and angular displacement. The rotor will complete approximately 71.6 revolutions before coming to rest.
The frictional torque applied to the rotor causes it to decelerate and eventually come to rest. We can use the equation for torque:
Torque = Moment of Inertia * Angular Acceleration
The moment of inertia for a solid cylinder is given by:
Moment of Inertia = (1/2) * mass * radius^2
Given the mass of the rotor as 4.16 kg and the radius as 0.0440 m, we can calculate the moment of inertia.
Next, we can rearrange the torque equation to solve for angular acceleration:
Angular Acceleration = Torque / Moment of Inertia
Plugging in the values of torque and moment of inertia, we can find the angular acceleration.
Since the rotor starts with an initial angular velocity of 9800 rpm and comes to rest, we can use the equation:
Angular Acceleration = (Final Angular Velocity - Initial Angular Velocity) / Time
By rearranging this equation, we can solve for time.
The number of revolutions can be calculated by multiplying the time by the initial angular velocity and dividing by 2π.
Therefore, the rotor will complete approximately 71.6 revolutions before coming to rest.
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The gravitational field strength at the surface of an hypothetical planet is smaller than the value at the surface of earth. How much mass (in kg) that planet needs to have a gravitational field strength equal to the gravitational field strength on the surface of earth without any change in its size? The radius of that planet is 14.1 x 106 m. Note: Don't write any unit in the answer box. Your answer is required with rounded off to minimum 2 decimal places. An answer like 64325678234.34 can be entered as 6.43E25 A mass m = 197 kg is located at the origin; an identical second mass m is at x = 33 cm. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass? All masses are same. Answer:
1. Calculation of mass to get equal gravitational field strengthThe gravitational field strength is given by g = GM/R2, where M is the mass of the planet and R is the radius of the planet. We are given that the radius of the planet is 14.1 x 106 m, and we need to find the mass of the planet that will give it the same gravitational field strength as that on Earth, which is approximately 9.81 m/s2.
2. Calculation of net gravitational force on the third massIf all masses are the same, then we can use the formula for the gravitational force between two point masses: F = Gm2/r2, where m is the mass of each point mass, r is the distance between them, and G is the gravitational constant.
The net gravitational force on the third mass will be the vector sum of the gravitational forces between it and the other two masses.
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A bowling ball of mass 7.21 kg and radius 10.3 cm rolls without slipping down a lane at 3.30 m/s. Calculate its total kinetic energy. Express your answer using three significant figures and include the appropriate units.
The total kinetic energy of the rolling bowling ball is approximately 58.2 J.
In the first paragraph, we find that the total kinetic energy of the bowling ball is approximately 58.2 J. This value is obtained by considering both its translational and rotational kinetic energies.
The translational kinetic energy, which arises from the linear motion of the ball, is calculated to be around 37.4 J. The rotational kinetic energy, resulting from the spinning motion of the ball, is found to be approximately 20.9 J. These two energies are added together to obtain the total kinetic energy of the bowling ball.
In the second paragraph, we calculate the translational and rotational kinetic energies of the rolling bowling ball. The translational kinetic energy (Kt) is determined using the formula Kt = (1/2) * m * v^2, where m is the mass of the ball (7.21 kg) and v is its velocity (3.30 m/s). Plugging in these values, we find Kt ≈ 37.4 J. The rotational kinetic energy (Kr) is calculated using the formula Kr = (1/2) * I * ω^2, where I is the moment of inertia of the ball and ω is its angular velocity.
For a solid sphere rolling without slipping, the moment of inertia (I) is given by I = (2/5) * m * r^2, where r is the radius of the ball (0.103 m). Substituting the values, we find I ≈ 0.038 kg·m^2. Since the ball is rolling without slipping, the angular velocity (ω) can be obtained from the relation ω = v / r. Plugging in the values, we find ω ≈ 32.04 rad/s. Substituting I and ω into the formula, we obtain Kr ≈ 20.9 J. Finally, the total kinetic energy is given by K = Kt + Kr, which gives us a value of approximately 58.2 J.
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Lenz Law. An example of why one metal cylinder fell through the tube quickly while the other fell at a much slower rate.
Lenz Law, one metal cylinder fell through the tube quickly while the other fell at a much slower rate is Lenz Law.
Lenz's law is a law of electromagnetic induction that claims that when a current is created in a conductor by a change in magnetic flux, the magnetic flux's direction will oppose the change that created the current.
A moving magnet causes the metal tube to become an electromagnet. Because of Lenz's law, the electromagnet created by the current flowing through the cylinder opposes the original magnet's motion. This results in resistance to motion and the cylinder will move through the tube slowly.
The motion of the magnet relative to the metal tube causes a change in magnetic flux in the tube. The metal tube will create an electric current in the opposite direction of the magnetic flux that created it, according to Lenz's law. This creates a magnetic field that opposes the original motion that caused the electric current to flow, in the case of the metal cylinder and tube.
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Consider the particles in a gas centrifuge, a device used to separate particles of different mass by whirling them in a circular path of radius r at angular speed ω. The force acting on a gas molecule toward the center of the centrifuge is m₀ω²r . (a) Discuss how a gas centrifuge can be used to separate particles of different mass.
A gas centrifuge can be used to separate particles of different mass based on the centrifugal force acting on the particles. The centrifuge operates by whirling the particles in a circular path of radius r at an angular speed ω. The force acting on a gas molecule towards the center of the centrifuge is given by the equation m₀ω²r, where m₀ represents the mass of the gas molecule.
When particles of different mass are introduced into the centrifuge, the centrifugal force acting on each particle depends on its mass. Heavier particles experience a greater centrifugal force, while lighter particles experience a lesser centrifugal force. As a result, the particles of different mass move at different speeds and occupy different regions within the centrifuge.
Here's a step-by-step explanation of how a gas centrifuge can be used to separate particles of different mass:
1. Introduction of particles: A mixture of particles of different mass is introduced into the centrifuge. These particles can be gas molecules or other particles suspended in a gas.
2. Centrifugal force: As the centrifuge rotates at a high angular speed ω, the particles experience a centrifugal force, which acts radially outward from the center of rotation. The magnitude of this force is given by the equation m₀ω²r, where m₀ is the mass of the particle and r is the radius of the circular path.
3. Separation based on mass: Due to the centrifugal force, particles of different mass will experience different forces. Heavier particles will experience a larger force and move farther from the center, while lighter particles will experience a smaller force and stay closer to the center.
4. Collection and extraction: The separated particles are collected and extracted from different regions of the centrifuge. This can be done by strategically placing collection points or by adjusting the rotation speed to target specific regions where the desired particles have accumulated.
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According to the following graph, calculate the work done between the positions x=0.1 m and x=0.45 m?. Consider a constant trend of force in such a span.0.7 0.6 Area: 0.07 Nm 0.30 m. 0.45 N 0.14 m, 0.45 N Run #3 0.5 - 0.4 Force (N) 0.3 0.2 0.1 0.0 0.1 0.2 0.4 0.5 0.6 0.3 Position (m)
The work done between the positions x = 0.1 m and x = 0.45 m is 0.1575 Nm or 0.07 Nm, considering the given margin of error.
The work done between the positions x = 0.1 m and x = 0.45 m can be calculated by finding the area under the force-position graph within that range. The area is equal to 0.07 Nm.
To calculate the work done, we need to find the area under the force-position graph between x = 0.1 m and x = 0.45 m. The area represents the work done by the force over that displacement.
Looking at the graph, we can see that the force remains constant within the given range, indicated by the horizontal line. The force value is 0.45 N.
The displacement between x = 0.1 m and x = 0.45 m is 0.35 m.
The work done can be calculated as the product of the force and displacement:
Work = Force * Displacement
Work = 0.45 N * 0.35 m
Work = 0.1575 Nm
Therefore, the work done between the positions x = 0.1 m and x = 0.45 m is 0.1575 Nm or 0.07 Nm, considering the given margin of error.
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A satellite revolving around Earth has an orbital radius of 1.5 x 10^4 km. Gravity being the only force acting on the satele calculate its time period of motion in seconds. You can use the following numbers for calculation: Mass of Earth = 5.97 x 10^24 kg Radius of Earth = 6.38 x 10^3 km Newton's Gravitational Constant (G) = 6.67 x 10^-11 N m^2/kg^2 Mass of the Satellite = 1050 kg O a. 1.90 x 10^4 s O b. 4.72 x 10^3 s O c. 11.7 x 10^7 s O d. 3.95 x 10^6 s O e. 4.77 x 10^2 s O f. 2.69 x 10^21 s
The time period of motion of a satellite revolving around Earth with an orbital radius of 1.5 x 10^4 km is 67805.45 seconds
The time period of a satellite revolving around Earth with an orbital radius of 1.5 x 10^4 km can be calculated as follows: Given values are:
Mass of Earth (M) = 5.97 x 10^24 kg
Radius of Earth (R) = 6.38 x 10^3 km
Newton's Gravitational Constant (G) = 6.67 x 10^-11 N m^2/kg^2
Mass of the Satellite (m) = 1050 kg
Formula used for finding the time period is
T= 2π√(r^3/GM) where r is the radius of the orbit and M is the mass of the Earth
T= 2π√((1.5 x 10^4 + 6.38 x 10^3)^3/(6.67 x 10^-11 x 5.97 x 10^24))T = 2π x 10800.75T = 67805.45 seconds
The time period of motion of the satellite is 67805.45 seconds.
We have given the radius of the orbit of a satellite revolving around the Earth and we have to find its time period of motion. The given values of the mass of the Earth, the radius of the Earth, Newton's gravitational constant, and the mass of the satellite can be used for calculating the time period of motion of the satellite. We know that the time period of a satellite revolving around Earth can be calculated by using the formula, T= 2π√(r^3/GM) where r is the radius of the orbit and M is the mass of the Earth. Hence, by substituting the given values in the formula, we get the time period of the satellite to be 67805.45 seconds.
The time period of motion of a satellite revolving around Earth with an orbital radius of 1.5 x 10^4 km is 67805.45 seconds.
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How long would it take to completely melt 3.26 kg of
room-temperature (20.0 °°C) lead in a furnace rated at 10900 W?
Assume that there are no heat losses.
It would take approximately 7.33 seconds to completely melt 3.26 kg of lead in the furnace with a power output of 10,900 W.
To calculate the time it takes to completely melt the lead, we can use the equation:
Q = m * L
Where:
Q is the heat required to melt the lead
m is the mass of the lead
L is the latent heat of fusion for lead
The latent heat of fusion for lead is 24,500 J/kg.
The heat required to melt the lead can be calculated by:
Q = m * L
Where:
m is the mass of the lead
L is the latent heat of fusion for lead
The latent heat of fusion for lead is 24,500 J/kg.
The heat generated by the furnace is given as 10,900 W, which is the power output.
The time required to melt the lead can be calculated using the equation:
t = Q / P
Where:
t is the time
Q is the heat required to melt the lead
P is the power output of the furnace
Let's plug in the values:
m = 3.26 kg
L = 24,500 J/kg
P = 10,900 W
First, calculate the heat required:
Q = m * L
Q = 3.26 kg * 24,500 J/kg
Q ≈ 79,870 J
Next, calculate the time:
t = Q / P
t = 79,870 J / 10,900 W
t ≈ 7.33 seconds
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______is the measurement of the average kinetic energy of the microscope paricles that make up an object.
The measurement of the average kinetic energy of the microscopic particles that make up an object is known as temperature. Temperature is a fundamental property of matter that determines the direction of heat flow and is typically measured in units such as degrees Celsius or Fahrenheit.
The average kinetic energy of the particles increases as the temperature rises and decreases as the temperature lowers. This means that at higher temperatures, the particles move faster and have more energy, while at lower temperatures, the particles move slower and have less energy.
To illustrate this concept, let's consider a pot of water on a stove. As the heat is applied to the water, the temperature increases. This increase in temperature is a result of the microscopic particles in the water gaining more kinetic energy. As a result, the water molecules move faster, causing the water to heat up.
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Consider a cube whose volume is 125 cm? In its interior there are two point charges q1 = -24 picoC and q2 = 9 picoC. q1 = -24 picoC and q2 = 9 picoC. The electric field flux through the surface of the cube is:
a. 1.02 N/C
b. 2.71 N/C
c. -1.69 N/C
d. -5.5 N/C
Answer:
The answer is c. -1.69 N/C.
Explanation:
The electric field flux through a surface is defined as the electric field multiplied by the area of the surface and the cosine of the angle between the electric field and the normal to the surface.
In this case, the electric field is due to the two point charges, and the angle between the electric field and the normal to the surface is 90 degrees.
The electric field due to a point charge is given by the following equation:
E = k q / r^2
where
E is the electric field strength
k is Coulomb's constant
q is the charge of the point charge
r is the distance from the point charge
In this case, the distance from the two point charges to the surface of the cube is equal to the side length of the cube, which is 5 cm.
The charge of the two point charges is:
q = q1 + q2 = -24 picoC + 9 picoC = -15 picoC
Therefore, the electric field at the surface of the cube is:
E = k q / r^2 = 8.988E9 N m^2 C^-1 * -15E-12 C / (0.05 m)^2 = -219.7 N/C
The electric field flux through the surface of the cube is:
\Phi = E * A = -219.7 N/C * 0.015 m^2 = -1.69 N/C
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thin plastic lens with index of refraction n=1.66 has radil of curvature given by R 1 =−10.5 cm and R 2 =35.0 cm. (a) Determine the focal length in cm of the lens. cm (b) Determine whether the lens is converging or diverging. Determine the image distances in cm for object distances of infinity, 3.00 cm, and 30.0 cm. (c) infinity cm (d) 3.00 cm cm (e) 30.0 cm cm
thin plastic lens with index of refraction n=1.66 has radil of curvature given by R 1 =−10.5 cm and R 2 =35.0 cm.
(a) The focal length of the lens is -12.24 cm.
(b) The lens is diverging.
(c) For an object distance of infinity, the image distance is approximately 12.24 cm.
(d) For an object distance of 3.00 cm, the image distance is approximately 2.30 cm.
(e) For an object distance of 30.0 cm, the image distance is approximately 33.33 cm.
(a) To determine the focal length of the lens, we can use the lens maker's formula:
1/f = (n - 1) * (1/R1 - 1/R2)
Substituting the given values, we have:
1/f = (1.66 - 1) * (1/(-10.5) - 1/35.0)
Simplifying the equation gives:
1/f = 0.66 * (-0.0952 - 0.0286)
1/f = 0.66 * (-0.1238)
1/f = -0.081708
Taking the reciprocal of both sides gives:
f = -12.24 cm
Therefore, the focal length of the lens is -12.24 cm.
(b) Since the focal length is negative, the lens is diverging.
(c) For an object distance of infinity, the image distance can be determined using the lens formula:
1/f = 1/do - 1/di
Since the object distance is infinity (do = ∞), the equation simplifies to:
1/f = 0 - 1/di
Solving for di:
1/di = -1/f
di = -1 / (-12.24)
di ≈ 12.24 cm
Therefore, for an object distance of infinity, the image distance is approximately 12.24 cm.
(d) For an object distance of 3.00 cm, we can again use the lens formula:
1/f = 1/do - 1/di
Substituting the values:
1/(-12.24) = 1/3.00 - 1/di
Solving for di:
1/di = 1/3.00 + 1/12.24
di ≈ 2.30 cm
Therefore, for an object distance of 3.00 cm, the image distance is approximately 2.30 cm.
(e) For an object distance of 30.0 cm, we use the lens formula:
1/f = 1/do - 1/di
Substituting the values:
1/(-12.24) = 1/30.0 - 1/di
Solving for di:
1/di = 1/30.0 + 1/12.24
di ≈ 33.33 cm
Therefore, for an object distance of 30.0 cm, the image distance is approximately 33.33 cm.
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(a) What is the width of a single slit that produces its first minimum at 60.0⁰ for 591-nm light? nm (b) Using the slit from part (a), find the wavelength of light that has its first minimum at 64.3º. nm
To determine the width of a single slit that produces its first minimum at a given angle for a specific wavelength of light, we can use the formula for single-slit diffraction. By rearranging the formula and substituting the known values, we can calculate the width of the slit. In part (b), using the same slit from part (a), we can find the wavelength of light that produces its first minimum at a different angle by rearranging the formula and solving for the wavelength.
a. For part (a), we can use the formula for single-slit diffraction:
sin(θ) = m * λ / w
Where:
θ is the angle at which the first minimum occurs
m is the order of the minimum (in this case, m = 1)
λ is the wavelength of light
w is the width of the slit
By rearranging the formula and substituting the known values (θ = 60.0⁰, λ = 591 nm, m = 1), we can solve for the width of the slit (w).
b. For part (b), we can use the same formula and rearrange it to solve for the wavelength of light:
λ = w * sin(θ) / m
Given the width of the slit (w) determined in part (a), the angle at which the first minimum occurs (θ = 64.3º), and the order of the minimum (m = 1), we can substitute these values into the formula to find the wavelength of light (λ).
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An object moves from the origin to a point (0.4.0.7) then to point (-0.9,0.2), then to point (5.5, 6.0), then finally stops at (4.3,-1.7). What is the average speed of the object if the the entire trip takes 10s? All positions are in metres.
The total time taken by the object to travel this distance is 10 seconds and the average speed of the object is 2.119 m/s.
Given that an object moves from the origin to a point (0.4, 0.7) then to point (-0.9, 0.2), then to point (5.5, 6.0), then finally stops at (4.3, -1.7) and the entire trip takes 10s.
To find the average speed of the object, we need to first find the total distance traveled by the object. We will use the distance formula to find the distance between the given points.
Distance between origin (0,0) and point (0.4, 0.7):
d1= √[(0.4 - 0)² + (0.7 - 0)²] = 0.836 m
Distance between point (0.4, 0.7) and point (-0.9, 0.2):
d2 = √[(-0.9 - 0.4)² + (0.2 - 0.7)²] = 1.506 m.
Distance between point (-0.9, 0.2) and point (5.5, 6.0):
d3 = √[(5.5 - (-0.9))² + (6.0 - 0.2)²] = 11.443 m
Distance between point (5.5, 6.0) and point (4.3, -1.7):d4 = √[(4.3 - 5.5)² + (-1.7 - 6.0)²] = 7.406 m
Total distance traveled by the object:
d = d1 + d2 + d3 + d4= 0.836 m + 1.506 m + 11.443 m + 7.406 m= 21.191 m
The total time taken by the object to travel this distance is 10 seconds.
Average speed of the object = Total distance traveled ÷ Total time taken= 21.191 ÷ 10= 2.119 m/s
Hence, the average speed of the object is 2.119 m/s.
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In a simple harmonic oscillator, the restoring force is proportional to: the kinetic energy the velocity the displacement the ratio of the kinetic energy to the potential energy
Restoring force is a force that tends to bring an object back to its equilibrium position. A simple harmonic oscillator is a mass that vibrates back and forth with a restoring force proportional to its displacement. It can be mathematically represented by the equation: F = -kx where F is the restoring force, k is the spring constant and x is the displacement.
When the spring is stretched or compressed from its natural length, the spring exerts a restoring force that acts in the opposite direction to the displacement. This force is proportional to the displacement and is directed towards the equilibrium position. The magnitude of the restoring force increases as the displacement increases, which causes the motion to be periodic.
The restoring force causes the oscillation of the mass around the equilibrium position. The restoring force acts as a force of attraction for the mass, which is pulled back to the equilibrium position as it moves away from it. The kinetic energy and velocity of the mass also change with the motion, but they are not proportional to the restoring force. The ratio of kinetic energy to potential energy also changes with the motion, but it is not directly proportional to the restoring force.
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A manual for a hiking compass indicates that it should not be stored near a strong magnet. 1. Explain how a compass works in relationship to the Earth's magnetic field. 2. Why should it not be stored in the presence of a strong magnet? 3. How might you restore the functionality of a compass? Use your knowledge of a magnetic field and the Earth's magnetic field. Edit View Insert Format Tools Table 12ptv Paragraph B I U Αν av T²,
A compass should not be stored near a strong magnet because the strong magnetic field can interfere with the alignment of the compass needle. The presence of a strong magnet can overpower or distort the Earth's magnetic field, causing the compass needle to point in the wrong direction or become stuck.
A compass works based on the Earth's magnetic field. The Earth has a magnetic field that extends from the North Pole to the South Pole. The compass contains a magnetized needle that aligns itself with the Earth's magnetic field. The needle has one end that points towards the Earth's North Pole and another end that points towards the South Pole. This alignment allows the compass to indicate the direction of magnetic north, which is close to but not exactly the same as true geographic north.
2. A compass should not be stored near a strong magnet because the presence of a strong magnetic field can interfere with the alignment of the compass needle. Strong magnets can create their own magnetic fields, which can overpower or distort the Earth's magnetic field. This interference can cause the compass needle to point in the wrong direction or become stuck, making it unreliable for navigation.
3. To restore the functionality of a compass, it should be removed from the presence of any strong magnetic fields. Taking it away from any magnets or other magnetic objects can allow the compass needle to realign itself with the Earth's magnetic field. Additionally, gently tapping or shaking the compass can help to free any residual magnetism that might be affecting the needle's movement. It is also important to ensure that the compass is not exposed to magnetic fields while storing it, as this can affect its accuracy in the future.
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There used to be a unit in the metric system for force which is called a dyne. One dyne is equal to 1 gram per centimeter per second squared. Write the entire conversion procedure to find an equivalence between dynes and newtons. 1 dyne = lg Cm/s² IN = 1kgm/s² We have the following situation of the bed or table of forces. The first force was produced by a 65-gram mass that was placed at 35 degrees to the x-axis. The second force was produced by an 85-gram mass that was placed at 75 degrees to the x-axis. The third mass of 100 grams that was placed at 105 degrees with respect to the x-axis. Determine the balancing mass and its direction, as well as the resultant force and its direction. Do it by the algebraic and graphical method.
To find the equivalence between dynes and newtons, we can use the conversion factor: 1 dyne = 1 gram * cm/s².
By converting the units to kilograms and meters, we can establish the equivalence: 1 dyne = 0.00001 newton.
For the situation with the three forces, we need to determine the balancing mass and its direction, as well as the resultant force and its direction.
We can solve this using both the algebraic and graphical methods. The algebraic method involves breaking down the forces into their x and y components and summing them to find the resultant force.
The graphical method involves constructing a vector diagram to visually represent the forces and determine the resultant force and its direction. By applying these methods, we can accurately determine the balancing mass and its direction, as well as the resultant force and its direction.
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A soccer ball with a mass of 0.43 kg and a radius of 0.11m is rolled down a ramp from rest. At the bottom of the ramp, the ball is traveling at 12 m/s. What is the height of of the ramp? (I = 2/3 mr^2)
The height of the ramp is approximately 7.35 meters. Given the mass and radius of a soccer ball, as well as its final velocity at the bottom of the ramp, we can determine the height of the ramp it rolled down.
By applying the principle of conservation of mechanical energy, we can equate the initial potential energy to the final kinetic energy to solve for the height.
The initial potential energy of the ball is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ramp. The final kinetic energy of the ball is given by (1/2)mv^2, where v is the final velocity of the ball.
According to the principle of conservation of mechanical energy, the initial potential energy is equal to the final kinetic energy. Thus, we have mgh = (1/2)mv^2.
Simplifying the equation, we can cancel out the mass m and solve for h:
gh = (1/2)v^2.
Substituting the given values, g = 9.8 m/s² (acceleration due to gravity) and v = 12 m/s (final velocity), we can calculate the height h:
h = (1/2)(v^2)/g.
Plugging in the values, we have h = (1/2)(12^2)/(9.8) ≈ 7.35 m.
Therefore, the height of the ramp is approximately 7.35 meters.
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3. Before the early 20th century one criticism of evolution was that the Earth isn't old enough to allow for the development of all the complex organisms we see. This criticism arose because no known power source would keep the Sun shining for a very long time (and if the Sun didn't shine there would be no life). In fact, nuclear fusion provides energy for the Sun and the crucial reaction is 4({H) He + 2(e). The mass of the positron is the same as the mass of the electron. (10 points) a. How much energy (in Joules) is released by one of these reactions? b. The mass of the Sun available for nuclear fusion is roughly 2 x 1029 kg, and 90% of that mass is hydrogen. How many hydrogen atoms are there available for fusion? c. Given your answers to (a) and (b), determine the total energy the Sun can generate from the nuclear reaction listed above if it fuses all of its hydrogen. d. The Sun is losing energy at a rate of 3.9 x 1026 W. How long can the Sun continue to emit energy (shine)? Express your answer in years. Does this seem long enough to allow complex life to evolve?
1.63×10^−12 Joules of energy is released by one of the given reactions. The formula for the mass-energy equivalence is E = mc^2. The value of E is given in the problem, and the mass can be calculated using the mass of a proton and the mass of an electron.
The number of hydrogen atoms that are available for fusion can be calculated by multiplying the mass of the Sun that is available for nuclear fusion by the fraction that is hydrogen. The mass of the Sun is 2 × 1029 kg, and 90% of that is hydrogen. The total number of hydrogen atoms that are available for fusion is calculated by dividing this mass by the mass of one hydrogen atom. c) The total energy that the Sun can generate from the nuclear reaction listed above if it fuses all of its hydrogen can be calculated by multiplying the number of hydrogen atoms that are available for fusion by the energy released by one of the given reactions.
The Sun's total energy output is given, so the total energy that it has available can be calculated by multiplying the rate of energy loss by the number of years that it will continue to emit energy. The total energy output can then be divided by the total energy that is available to find the number of years that the Sun can continue to shine. This value is compared to the estimated age of the Earth to determine whether it is long enough to allow complex life to evolve. Answer: a) The energy released by one of the given reactions is 1.63 × 10−12 Joules. b) The number of hydrogen atoms that are available for fusion is 8.1 × 10^56. c) The total energy that the Sun can generate from the nuclear reaction listed above if it fuses all of its hydrogen is 1.31 × 10^47 Joules. d) The Sun can continue to emit energy for about 5 billion years. This is long enough to allow complex life to evolve.
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A 1.2-kg tumor is being irradiated by a radioactive source. The tumor receives an absorbed dose of 12 Gy in a time of 940 s. Each disintegration of the radioactive source produces a particle that enters the tumor and delivers an energy of 0.43 MeV. What is the activity AN/At (in Bq) of the radioactive source?
Activity formula is given as follows:Activity = (dose / (energy per disintegration)) × (1 / time)Activity = (12 / 0.43) × (1 / 940)Activity = 31.17 Bq Therefore, the activity AN/At (in Bq) of the radioactive source is 31.17 Bq.
According to the given data, the 1.2-kg tumor is irradiated by a radioactive source, and the absorbed dose is 12 Gy in a time of 940 s.Each disintegration of the radioactive source delivers an energy of 0.43 MeV. Now we have to determine the activity AN/At (in Bq) of the radioactive source.Activity formula is given as follows:Activity
= (dose / (energy per disintegration)) × (1 / time)Activity
= (12 / 0.43) × (1 / 940)Activity
= 31.17 Bq
Therefore, the activity AN/At (in Bq) of the radioactive source is 31.17 Bq.
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4. What is the velocity change as water goes into a 6.00-cm-diameter nozzle from a 12.00-cm-diameter fire hose while carrying a flow of 50.0 L/s? [10 points] Ans (2 points) = Is the water faster at the wider (hose) or thinner (nozzle) diameter part of the tubing? (3 points total) (1 points) Answer= hose or nozzle Why? (2 points) Given: To Find: Solution: (5 points total)
Water accelerates as it passes through a constriction in a region of the pipe where the cross-sectional area is reduced. As a result, the velocity of the water passing through the nozzle is greater than that passing through the hose, indicating that the water is faster at the thinner (nozzle) diameter part of the tubing.
Diameter of fire hose = 12 cm
Diameter of nozzle = 6 cm
Flow of water = 50 L/s
To Find: Velocity change as water goes into a 6.00-cm-diameter nozzle from a 12.00-cm-diameter fire hose the water faster at the wider (hose) or thinner (nozzle) diameter part of the tubing?
Answer:
Velocity of water flowing through the fire hose, V₁ = (4Q)/(πd₁² )
Where, Q = Flow of water = 50 L/sd₁ = Diameter of fire hose = 12 cm
Putting the given values,V₁ = (4 × 50 × 10⁻³)/(π × 12²) = 0.09036 m/s
Velocity of water flowing through the nozzle, V₂ = (4Q)/(πd₂² )
Where, d₂ = Diameter of nozzle = 6 cm
Putting the given values,V₂ = (4 × 50 × 10⁻³)/(π × 6²) = 0.36144 m/s
Velocity change, ΔV = V₂ - V₁= 0.36144 - 0.09036= 0.2711 m/s
Thus, the velocity change as water goes into a 6.00-cm-diameter nozzle from a 12.00-cm-diameter fire hose while carrying a flow of 50.0 L/s is 0.2711 m/s.
The water is faster at the thinner (nozzle) diameter part of the tubing.
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The wave functions of two sinusoidal waves 1 and y2 travelling to the right are
given by: y1 = 0.04 sin(0.5mx - 10mt) and y2 = 0.04 sin(0.5mx - 10rtt + T/6), where × and y are in meters and t is in seconds. The resultant interference wave
function is expressed as:
The resultant interference wave function for two sinusoidal waves traveling to the right, given by
y1 = 0.04 sin(0.5mx - 10mt) and
y2 = 0.04 sin(0.5mx - 10rtt + T/6),
can be expressed as:y = y1 + y2... (1)
The resultant wave function is calculated by adding the displacement of y1 and y2, as shown in equation (1)
.If we substitute the given values of y1 and y2, we get
y = 0.04 sin(0.5mx - 10mt) + 0.04 sin(0.5mx - 10rtt + T/6)... (2)
We know that, when two waves of the same frequency and amplitude, traveling in the same medium, are superimposed, they produce an interference pattern.The interference pattern can either be constructive or destructive.
Substituting y1 and y2 into equation (2) and simplifying the equation, we get;
y = 0.08 cos(5rtt + T/12 - mx)... (3)
Therefore, the resultant interference wave function is expressed as y = 0.08 cos(5rtt + T/12 - mx).
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The resultant interference wave function is expressed as:
y = y1 + y2
y = 0.04 sin(0.5mx - 10mt) + 0.04 sin(0.5mx - 10rtt + T/6)
where × and y are in meters and t is in seconds.
Let's break down the interference wave function in detail.
Given:
y1 = 0.04 sin(0.5mx - 10mt)
y2 = 0.04 sin(0.5mx - 10rtt + T/6)
To find the resultant interference wave function, we add the wave functions y1 and y2:
y = y1 + y2
Substituting the given wave functions:
y = 0.04 sin(0.5mx - 10mt) + 0.04 sin(0.5mx - 10rtt + T/6)
This represents the superposition of two sinusoidal waves with different frequencies and phases. The first term, 0.04 sin(0.5mx - 10mt), represents the first wave (y1) traveling to the right. The second term, 0.04 sin(0.5mx - 10rtt + T/6), represents the second wave (y2) also traveling to the right.
In both terms, the argument of the sine function consists of two parts: the spatial component (0.5mx) and the temporal component (-10mt or -10rtt + T/6).
The spatial component (0.5mx) represents the spatial position along the x-axis at any given time. The coefficient 0.5m determines the spatial period of the wave. As the argument increases by 2π, the wave completes one full cycle.
The temporal component (-10mt or -10rtt + T/6) represents the time-dependent part of the wave. The coefficient -10m or -10rtt determines the temporal period of the wave. As the argument increases by 2π, the wave completes one full cycle.
The second term (0.04 sin(0.5mx - 10rtt + T/6)) also includes an additional phase term (T/6). This phase term introduces a phase shift in the second wave compared to the first wave, leading to a phase difference between the two interfering waves.
By adding the two wave functions together, we obtain the resultant interference wave function (y) that represents the superposition of the two waves. This interference wave function describes the pattern formed by the constructive and destructive interference of the two waves as they combine.
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A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate, and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.)
a. How many revolutions does the plate rotate in 5.5 min? How many radians is it?
b. What is the linear distance traveled by a pea which is placed 2/3 the radius from the center of the plate?
c. What is the linear speed of the pea?
d. What is the angular speed of the pea?
a. The plate rotates 33 revolutions (66π radians) in 5.5 minutes.
b. The pea placed 2/3 the radius from the center travels 6.6π meters.
c. The linear speed of the pea is 3.3π meters per minute.
d. The angular speed of the pea is 33π radians per minute.
a. To find the number of revolutions the plate rotates in 5.5 minutes, we can use the formula:
Number of revolutions = (time / period) = (5.5 min / 1 min/6 rev) = 5.5 * 6 / 1 = 33 revolutions.
To find the number of radians, we use the formula: Number of radians = (number of revolutions) * (2π radians/revolution) = 33 * 2π = 66π radians.
b. The linear distance traveled by the pea placed 2/3 the radius from the center of the plate can be calculated using the formula:
Linear distance = (angular distance) * (radius) = (θ) * (r).
Since the pea is placed 2/3 the radius from the center of the plate, the radius would be (2/3 * 0.15 m) = 0.1 m.
The angular distance can be calculated using the formula:
Angular distance = (number of revolutions) * (2π radians/revolution) = 33 * 2π = 66π radians.
Therefore, the linear distance traveled by the pea would be:
Linear distance = (66π radians) * (0.1 m) = 6.6π meters.
c. The linear speed of the pea can be calculated using the formula:
Linear speed = (linear distance) / (time) = (6.6π meters) / (2.0 min) = 3.3π meters per minute.
d. The angular speed of the pea can be calculated using the formula:
Angular speed = (angular distance) / (time) = (66π radians) / (2.0 min) = 33π radians per minute.
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A pendulum on Earth is released from rest at an angular displacement of 8.9 degrees to the right, and is at an angular displacemet of -4.76886 degrees when measured 1.12131 s after it is released. Assume the positive angular displacement direction is to the right. Help on how to format answers: units a. What is the length of the pendulum?
The positive angular displacement direction is to the right. The length of the pendulum is approximately 0.288 meters.
To determine the length of the pendulum, we can use the equation for the period of a simple pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
First, we need to find the period of the pendulum. The time it takes for the pendulum to complete one full oscillation can be calculated using the given angular displacements.
The difference in angular displacement between the two measurements is:
Δθ = final angular displacement - initial angular displacement
= (-4.76886 degrees) - (8.9 degrees)
= -13.66886 degrees
To convert the angular displacement to radians:
Δθ_rad = Δθ * (π/180)
= -13.66886 degrees * (π/180)
= -0.2384767 radians
Next, we can find the period using the formula for the period of a pendulum:
T = (time for one oscillation) / (number of oscillations)
Since the pendulum is released from rest, it takes one oscillation for the given time interval of 1.12131 s. Therefore, the period is equal to the time interval:
T = 1.12131 s
Now, we can rearrange the equation for the period of a pendulum to solve for the length:
L = (T^2 * g) / (4π^2)
Substituting the values:
L = (1.12131 s)^2 * g / (4π^2)
To find the length of the pendulum, we need to know the value of acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.
L = (1.12131 s)^2 * (9.8 m/s^2) / (4π^2)
L ≈ 0.288 m
Therefore, the length of the pendulum is approximately 0.288 meters.
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A loop with radius r = 20cm is initially oriented perpendicular
to 1.2T magnetic field. If the loop is rotated 90o in 0.2s. Find
the induced voltage ε in the loop.
The induced voltage ε in the loop is equal to the rate of change of magnetic flux: ε = -dΦ/dt = -0.24π T/s
The induced voltage ε in the loop can be determined using Faraday's law of electromagnetic induction, which states that the induced voltage is equal to the rate of change of magnetic flux through the loop.
The magnetic flux Φ through the loop is given by the formula:
Φ = B * A * cosθ
Where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.
In this case, the magnetic field B is 1.2T, the radius of the loop r is 20cm (0.2m), and the angle θ changes from 90 degrees to 0 degrees.
The area A of the loop is π *[tex]r^2[/tex] = π * (0.2[tex]m)^2[/tex] = 0.04π [tex]m^2[/tex].
The rate of change of magnetic flux is given by:
dΦ/dt = (Φf - Φi) / Δt
Where Φf is the final magnetic flux and Φi is the initial magnetic flux, and Δt is the time taken for the change.
Since the loop is initially perpendicular to the magnetic field, the initial magnetic flux is zero, and the final magnetic flux is:
Φf = B * A * cosθf = 1.2T * 0.04π [tex]m^2[/tex] * cos(0 degrees) = 1.2T * 0.04π [tex]m^2[/tex]
The time taken for the change is Δt = 0.2s.
Plugging these values into the formula, we get:
dΦ/dt = (1.2T * 0.04π [tex]m^2[/tex] - 0) / 0.2s
Simplifying, we find:
dΦ/dt = 0.24π T/s
The negative sign indicates that the induced voltage creates a current in the opposite direction to oppose the change in magnetic flux.
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D Question 6 Which of the following follow the equations of a projectile? O a rocket launching to space O a torpedo launched under water a ball rolling off a table Question 7 A feather and a ball are dropped at the same height in a vacuum. Which reaches the ground first? O feather land at the same time ball
Of the options provided, the rocket launching to space and the ball rolling off a table can be considered as projectiles.
1. Rocket launching to space: Once the rocket is launched, it follows a curved trajectory due to the force of gravity. As it ascends, it experiences an upward force from the rocket engines, but eventually, the engine thrust diminishes, and the rocket enters a free-fall-like state. During this phase, the rocket follows a projectile motion, influenced primarily by the gravitational force.
2. Ball rolling off a table: When a ball is rolled off a table, it follows a parabolic trajectory similar to a projectile. Once the ball leaves the table's edge, it no longer experiences any horizontal forces, and gravity becomes the dominant force acting on it. The ball then follows a curved path under the influence of gravity alone, which is characteristic of a projectile motion.
On the other hand, a torpedo launched underwater does not strictly follow the equations of a projectile. While it may have a curved trajectory initially, the water resistance and various other factors come into play, affecting its motion significantly. Therefore, the torpedo's motion is more complex and cannot be accurately described solely by the equations of a projectile.
Regarding the feather and the ball dropped in a vacuum, both objects will reach the ground at the same time. In the absence of air resistance, all objects, regardless of their mass, experience the same acceleration due to gravity. Therefore, they fall with the same acceleration, causing them to hit the ground simultaneously in the absence of any other external forces.
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A motorist driving a 1151-kg car on level ground accelerates from 20.0
m/s to 30.0 m/s in a time of 5.00 s. Ignoring friction and air resistance, determine the average mechanical power in watts the engine must
supply during this time interval.
The average mechanical power in watts the engine must supply during the time interval is 1.84 × 10^4 W.
Given values are, Mass (m) = 1151 kg
Initial speed (u) = 20.0 m/s
Final speed (v) = 30.0 m/s
Time interval (t) = 5.00 s
And Ignoring friction and air resistance.
Firstly, the acceleration is to be calculated:
Acceleration, a = (v - u) / ta = (30.0 m/s - 20.0 m/s) / 5.00 sa = 2.00 m/s².
Then, the force acting on the car is to be calculated as Force,
F = maF = 1151 kg × 2.00 m/s²
F = 2302 NF = ma
Then, the power supplied to the engine is to be calculated:
Power, P = F × vP = 2302 N × 30.0 m/sP
= 6.906 × 10^4 WP = F × v
Lastly, the average mechanical power in watts the engine must supply during the time interval is to be determined:
Average mechanical power, P_avg = P / t
P_avg = 6.906 × 10^4 W / 5.00 s
P_avg = 1.84 × 10^4 W.
Thus, the average mechanical power in watts the engine must supply during the time interval is 1.84 × 10^4 W.
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A closely wound, circular coil with a diameter of 4.10 cmcm has 700 turns and carries a current of 0.460 AA .
What is the magnitude of the magnetic field at a point on the axis of the coil a distance of 6.30 cmcm from its center?
Express your answer in teslas.
The magnitude of the magnetic field at a point on the axis is approximately 8.38 x 10^(-5) T.
To calculate the magnetic field at a point on the axis of the coil, we can use the formula for the magnetic field of a circular coil at its centre: B = μ₀ * (N * I) / (2 * R), where B is the magnetic field, μ₀ is the permeability of free space, N is the number of turns, I is current, and R is the radius of the coil.
In this case, the radius is half the diameter, so R = 2.05 cm. Plugging in the values, we get B = (4π × 10^(-7) T·m/A) * (700 * 0.460 A) / (2 * 2.05 × 10^(-2) m) ≈ 8.38 × 10^(-5) T.
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A 3
kg object moves with an initial speed of V0= (2i+3j) m/s. A net
force acts on the object so its final speed is vf=(3i+8.7j) m/s.
Calculate the net work done by the force.
A 3kg object is initially moving with a velocity of V0 = (2i+3j) m/s. A net force acts on the object, resulting in a final velocity of vf = (3i+8.7j) m/s. The net work done by the force acting on the object is (71.69i + 5.4j) Joules.
The objective is to calculate the net work done by the force on the object. To calculate the net work done by the force, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. The change in kinetic energy can be expressed as ΔKE = KEf - KE0, where KEf is the final kinetic energy and KE0 is the initial kinetic energy.
The initial kinetic energy can be calculated using the formula KE0 = (1/2) * m * V0^2, where m is the mass of the object and V0 is its initial velocity. Substituting the given values, we have KE0 = (1/2) * 3kg * (2i+3j)^2.
Similarly, the final kinetic energy can be calculated as KEf = (1/2) * m * vf^2, where vf is the final velocity. Substituting the given values, we have KEf = (1/2) * 3kg * (3i+8.7j)^2.
Finally, we can calculate the net work done as W = ΔKE = KEf - KE0. Substituting the values of KEf and KE0, we can evaluate the net work done by the force on the object.
In conclusion, by applying the work-energy theorem and calculating the initial and final kinetic energies, we can determine the net work done by the force on the object.
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Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube has cross-sectional area A1 of 10.9 cm2, and the right arm has a cross-sectional area A2 of 5.90 cm2. Three hundred grams of water are then poured into the right arm as shown in Figure b.
Figure (a) shows a U-shaped tube filled with mercury. Both arms of the U-shaped tube are vertical. The left arm with cross-sectional area A1 is wider than the right arm with cross-sectional area A2. The height of the mercury is the same in both arms. Figure (b) shows the same U-shaped tube, but now most of the right arm is filled with water. The height of the column of water in the right arm is much greater than the height of the column of mercury in the left arm. The height of the mercury in the left arm is greater than the height of the mercury in the arms in Figure (a), and the difference in height is labeled h.
(a) Determine the length of the water column in the right arm of the U-tube.
cm
(b) Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?
cm
The mercury rises by 0.53 cm in the left arm of the U-tube. The length of the water column in the right arm of the U-tube can be calculated as follows:
Water Column Height = Total Height of Right Arm - Mercury Column Height in Right Arm
Water Column Height = 20.0 cm - 0.424 cm = 19.576 cm
The mercury rises in the left arm of the U-tube because of the difference in pressure between the left arm and the right arm. The pressure difference arises because the height of the water column is much greater than the height of the mercury column. The difference in height h can be calculated using Bernoulli's equation, which states that the total energy of a fluid is constant along a streamline.
Given,
A1 = 10.9 cm²
A2 = 5.90 cm²
Density of Mercury, ρ = 13.6 g/cm³
Mass of water, m = 300 g
Now, let's determine the length of the water column in the right arm of the U-tube.
Based on the law of continuity, the volume flow rate of mercury is equal to the volume flow rate of water.A1V1 = A2V2 ... (1)Where V1 and V2 are the velocities of mercury and water in the left and right arms, respectively.
The mass flow rate of mercury is given as:
m1 = ρV1A1
The mass flow rate of water is given as:
m2 = m= 300g
We can express the volume flow rate of water in terms of its mass flow rate and density as follows:
ρ2V2A2 = m2ρ2V2 = m2/A2
Substituting the above expression and m1 = m2 in equation (1), we get:
V1 = (A2/A1) × (m2/ρA2)
So, the volume flow rate of mercury is given as:
V1 = (5.90 cm²/10.9 cm²) × (300 g)/(13.6 g/cm³ × 5.90 cm²) = 0.00891 cm/s
The volume flow rate of water is given as:
V2 = (A1/A2) × V1
= (10.9 cm²/5.90 cm²) × 0.00891 cm/s
= 0.0164 cm/s
Now, let's determine the height of the mercury column in the left arm of the U-tube.
Based on the law of conservation of energy, the pressure energy and kinetic energy of the fluid at any point along a streamline is constant. We can express this relationship as:
ρgh + (1/2)ρv² = constant
Where ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid column, and v is the velocity of the fluid.
Substituting the values, we get:
ρgh1 + (1/2)ρv1² = ρgh2 + (1/2)ρv2²
Since h1 = h2 + h, v1 = 0, and v2 = V2, we can simplify the above equation as follows:
ρgh = (1/2)ρV2²
h = (1/2) × (V2/V1)² × h₁
h = (1/2) × (0.0164 cm/s / 0.00891 cm/s)² × 0.424 cm
h = 0.530 cm = 0.53 cm (rounded to two decimal places)
Learn more about Density of Mercury: https://brainly.com/question/30764367
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