At t = 1.5 s, the changing current in the solenoid induces an EMF (electromotive force) of approximately 7.91 V in the coaxial coil.
To calculate the induced EMF in the coil, we need to determine the magnetic flux through the coil and then apply Faraday's law of electromagnetic induction.
1. Magnetic flux through the coil:
The magnetic flux through the coil is given by the equation Φ = B · A · N, where B is the magnetic field, A is the area of the coil, and N is the number of turns.
The magnetic field inside a solenoid is given by the equation B = μ₀ · n · I, where μ₀ is the permeability of free space, n is the number of turns per meter, and I is the current flowing through the solenoid.
Substituting the given values, the magnetic field inside the solenoid is B = (4π × 10⁻⁷ T·m/A) · (4000 turns/m) · [50 (1e^(-1.6 × 1.5)) A].
The area of the coil is A = π · R², where R is the radius of the coil.
2. EMF induced in the coil:
According to Faraday's law of electromagnetic induction, the induced EMF in the coil is given by the equation ε = -dΦ/dt, where ε is the induced EMF and dΦ/dt is the rate of change of magnetic flux.
To find the rate of change of magnetic flux, we need to differentiate the magnetic flux equation with respect to time. Since the magnetic field inside the solenoid is changing with time, we also need to consider the time derivative of the magnetic field.
Finally, substitute the values at t = 1.5 s into the derived equation to calculate the induced EMF in the coil.
By following these steps, we find that at t = 1.5 s, the changing current in the solenoid induces an EMF of approximately 7.91 V in the coaxial coil.
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A wheel undergoing MCUV rotates with an angular speed of 50 rad/s at t = 0 s and the magnitude of its angular acceleration is α = 5 rad/s^2. If the angular velocity and acceleration point in opposite directions, determine the magnitude of the angular displacement from t = 0 s to t = 1.1 s.
- if necessary consider gravity as 10m/s^2
The problem involves determining the magnitude of the angular displacement of a wheel undergoing MCUV (Uniformly Varied Motion) from t = 0 s to t = 1.1 s. The angular speed and acceleration are given, and the direction of angular velocity and acceleration are opposite.
The angular displacement of an object undergoing MCUV can be calculated using the equation θ = ω₀t + (1/2)αt², where θ is the angular displacement, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time interval.
Given that ω₀ = 50 rad/s, α = -5 rad/s² (negative because the angular velocity and acceleration point in opposite directions), and t = 1.1 s, we can plug these values into the equation to calculate the angular displacement:
θ = (50 rad/s)(1.1 s) + (1/2)(-5 rad/s²)(1.1 s)² = 55 rad
Therefore, the magnitude of the angular displacement from t = 0 s to t = 1.1 s is 55 rad. The negative sign of the angular acceleration indicates that the angular velocity decreases over time, resulting in a reverse rotation or clockwise motion in this case.
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A load is suspended from a steel wire with a radius of 1 mm. The load extends the wire the same amount as heating by 20°С. Find the weight of the load
The weight of the load is 0.128 kg.
Radius of the wire = 1 mm
Extension in the wire = Heating by 20°С
Weight of the load = ?
Formula used: Young's Modulus (Y) = Stress / Strain
When a wire is extended by force F, the strain is given as,
Strain = extension / original length
Where the original length is the length of the wire before loading and extension is the increase in length of the wire after loading.
Suppose the cross-sectional area of the wire be A. If T be the tensile force in the wire then Stress = T/A.
Now, according to Young's modulus formula,
Y = Stress / Strain
Solving the above expression for F, we get,
F = YAΔL/L
Where F is the force applied
YA is the Young's modulus of the material
ΔL is the change in length
L is the original length of the material
Y for steel wire is 2.0 × 1011 N/m2Change in length, ΔL = Original Length * Strain
Where strain is the increase in length per unit length
Original Length = 2 * Radius
= 2 * 1 mm
= 2 × 10⁻³ m
Strain = Change in length / Original length
Let x be the weight of the load, the weight of the load acting downwards = Force (F) acting upwards
F = xN
By equating both the forces and solving for the unknown variable x, we can obtain the weight of the load.
Solution:
F = YAΔL/L
F = (2.0 × 1011 N/m²) * π (1 × 10⁻³ m)² * (20°C) * (2 × 10⁻³ m) / 2 × 10⁻³ m
F = 1.256 N
f = mg
x = F/g
= 1.256 N / 9.8 m/s²
= 0.128 kg
Therefore, the weight of the load is 0.128 kg.
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A hailstone of mass 5.80 10-4 kg falls
through the air and experiences a net force given by the following
formula.
(a) If C = 2.85 10-5 kg/m,
calculate the terminal speed of the hailstone.
_________
The terminal velocity of the hailstone is 11.8 m/s.
The formula given is F_net = CρAg + mg, where F_net is the net force, C is the drag coefficient, ρ is the density of the fluid, A is the projected area of the object, g is the acceleration due to gravity, and m is the mass of the object.
Now, we can determine the terminal speed of the hailstone.
(a) If C = 2.85 × 10⁻⁵ kg/m, calculate the terminal speed of the hailstone. We can use the formula:
v_terminal = (2mg / ρCπr²)¹/²
where v_terminal is the terminal velocity, m is the mass of the hailstone, ρ is the density of air, C is the drag coefficient, and r is the radius of the hailstone.
v_terminal = (2mg / ρCπr²)¹/²
= [2(5.80 × 10⁻⁴ kg)(9.8 m/s²)] / [1.20 kg/m³ × (2.85 × 10⁻⁵ kg/m) × π (0.5 × 1.25 × 10⁻³ m)²]¹/²
= 11.8 m/s
Therefore, the terminal velocity of the hailstone is 11.8 m/s.
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Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of -8.80 μC/m², and sheet B, which is to the right of A, carries a uniform charge density of -11.6 μC/m². Assume that the sheets are large enough to be treated as infinite.
Part A: Find the magnitude of the net electric field (in units of electric field ) these sheets produce at a point 4.00 cm to the right of sheet A.
Part B: Find the direction of this net electric field.
Part B: Find the magnitude of the net electric field (in units of electric field) these sheets produce at a point 4.00 cm to the left of sheet A. Find the direction of this net electric field.
Part C: Find the magnitude of the net electric field (in units of electric field) these sheets produce at a point 4.00 cm to the right of sheet B. Find the direction of this net electric field.
For a point located 4.00 cm to the right of sheet A, the magnitude of the net electric field produced by the two sheets is approximately 3.07 × 10^4 N/C directed to the right. For a point located 4.00 cm to the left of sheet A, the magnitude of the net electric field is approximately 3.41 × 10^4 N/C directed to the left. For a point located 4.00 cm to the right of sheet B, the magnitude of the net electric field is approximately 2.28 × 10^4 N/C directed to the right.
To find the net electric field produced by the two sheets, we can calculate the electric field due to each sheet individually and then combine them. The electric field due to an infinite sheet of charge is given by the equation E = σ / (2ε₀), where E is the electric field, σ is the surface charge density, and ε₀ is the permittivity of free space.
For Part A, the electric field due to sheet A is E₁ = σ₁ / (2ε₀), where σ₁ = -8.80 μC/m². The electric field due to sheet B is E₂ = σ₂ / (2ε₀), where σ₂ = -11.6 μC/m². Since the electric fields of the two sheets are in the same direction, we can simply add them together. Therefore, the net electric field at a point 4.00 cm to the right of sheet A is E = E₁ + E₂.
For Part B, the magnitude of the net electric field can be calculated using the same method as Part A, but now the point of interest is 4.00 cm to the left of sheet A. Since the electric fields of the two sheets are in opposite directions, we subtract the electric field due to sheet B from the electric field due to sheet A to find the net electric field.
For Part C, we calculate the electric field due to sheet B at a point 4.00 cm to the right of sheet B using the equation E₂ = σ₂ / (2ε₀). Since sheet A is not involved in this calculation, the net electric field is simply equal to the electric field due to sheet B.
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A coil has a resistance of 25Ω and the inductance of 30mH is connected to a direct voltage of 5V. Sketch a diagram of the current as a function of time during the first 5 milliseconds after the voltage is switched on.
Answer:
A coil with a resistance of 25 ohms and an inductance of 30 millihenries is connected to a direct voltage of 5 volts.
The current will increase linearly for the first 0.75 milliseconds, and then reach a maximum value of 0.2 amperes. The current will then decrease exponentially.
Explanation:
A coil with a resistance of 25 ohms and an inductance of 30 millihenries is connected to a direct voltage of 5 volts.
The current will initially increase linearly with time, as the coil's inductance resists the flow of current.
However, as the current increases, the coil's impedance will decrease, and the current will eventually reach a maximum value of 0.2 amperes. The current will then decrease exponentially, with a time constant of 0.75 milliseconds.
The following graph shows the current as a function of time during the first 5 milliseconds after the voltage is switched on:
Current (A)
0.5
0.4
0.3
0.2
0.1
0
Time (ms)
0
1
2
3
4
5
The graph shows that the current increases linearly for the first 0.75 milliseconds, and then reaches a maximum value of 0.2 amperes. The current then decreases exponentially, with a time constant of 0.75 milliseconds.
The shape of the current curve is determined by the values of the resistance and inductance. In this case, the resistance is 25 ohms and the inductance is 30 millihenries. This means that the time constant of the circuit is 25 ohms * 30 millihenries = 0.75 milliseconds.
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8)The electric field in a sine wave has a peak value of 32.6 mV/m. Calculate the magnitude of the Poynting vector in this case.
The Poynting vector is the power density of an electromagnetic field.
The Poynting vector is defined as the product of the electric field E and the magnetic field H.
The Poynting vector in this case can be calculated by:
S = E × H
where E is the electric field and H is the magnetic field.
E/B = c
where c is the speed of light and B is the magnetic field.
[tex]E/B = c⇒ B = E/c⇒ B = (32.6 × 10⁻³)/(3 × 10⁸) = 1.087 × 10⁻¹¹[/tex]
The magnitude of the magnetic field H is then:
B = μH
where μ is the magnetic permeability of free space, which has a value of [tex]4π × 10⁻⁷ N/A².[/tex]
[tex]1.087 × 10⁻¹¹/(4π × 10⁻⁷) = 8.690H = 5 × 10⁻⁷[/tex]
The Poynting vector is then:
[tex]S = E × H = (32.6 × 10⁻³) × (8.6905 × 10⁻⁷) = 2.832 × 10⁻⁹ W/m²[/tex]
The magnitude of the Poynting vector in this case is 2.832 × 10⁻⁹ W/m².
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7. () An EM wave has an electric field given by E= (200 V/m) [sin ((0.5m-¹)z-(5 x 10°rad/s)t)] 3. Find a) Find the wavelength of the wave. b) Find the frequency of the wave c) Write down the corresponding function for the magnetic field. 8. () A beam of light strikes the surface of glass (n = 1.46) at an angle of 70° with respect to the normal. Find the angle of refraction inside the glass. Take the index of refraction of air n₁ = 1. 9. () A transformer has 350 turns in its primary coil and 400 turns in its secondary coil. If a voltage of 110 V is applied to its primary, find the voltage in its secondary.
Find the wavelength of the waveThe wavelength of the EM wave can be calculated from the equation λ = v/f, where λ is the wavelength, v is the speed of light, and f is the frequency.Therefore, the voltage in the secondary is 126V.
.λ = c/f
where c is the speed of light= 3x108/5x1010
= 6x10-3 m
The frequency of the EM wave is given as f = (5 x 10¹⁰ rad/s)/(2π)
= 2.5 x 10⁹ Hz.
E/B = c,
where E is the electric field, B is the magnetic field, and c is the speed of light. So,
B = E/c
=200/3x108 sin ((0.5m-¹)z-(5 x 10°rad/s)t)]
A beam of light strikes the surface of glass at an angle of 70° with respect to the normal.
index of refraction of air n₁ = 1.
Using Snell's law of refraction
: n1 sin θ1 = n2 sin θ2
Where n1 is the index of refraction of the medium of incidence, θ1 is the angle of incidence, n2 is the index of refraction of the refracting medium, and θ2 is the angle of refraction.n
₁sinθ1 = n₂sinθ2sinθ2
= n₁/n₂sinθ2
= 1/1.46 x sin70°sinθ2
= 0.4624θ2
= sin-1(0.4624)θ2
= 28.3°
Therefore, the angle of refraction inside the glass is 28.3°.9.A transformer has 350 turns in its primary coil and 400 turns in its secondary coil. If a voltage of 110 V is applied to its primary, find the voltage in its secondary.The voltage ratio of a transformer is given by the formula
:Ns / Np = Vs / V
where Ns and Np are the numbers of turns in the secondary an
primary coils respectively, and Vs and Vp are the voltages across the secondary and primary coils respectively
.So,Vs = (Ns/Np) * VpVs
= (400/350) * 110Vs
= 126V
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: Suppose 45 cm of wire is experiencing a magnetic force of 0.55 N. 50% Part (a) What is the angle in degrees between the wire and the 1.25 T field if it is carrying a 6.5 A current?
To find the angle between the wire and the magnetic field, we can use the formula for the magnetic force on a current-carrying wire:
F = BILsinθ
Where:
F = Magnetic force
B = Magnetic field strength
I = Current
L = Length of the wire
θ = Angle between the wire and the magnetic field
We are given:
F = 0.55 N
B = 1.25 T
I = 6.5 A
L = 45 cm = 0.45 m
Let's rearrange the formula to solve for θ:
θ = sin^(-1)(F / (BIL))
Substituting the given values:
θ = sin^(-1)(0.55 N / (1.25 T * 6.5 A * 0.45 m))
Now we can calculate θ:
θ = sin^(-1)(0.55 / (1.25 * 6.5 * 0.45))
Using a calculator, we find:
θ ≈ sin^(-1)(0.0558)
θ ≈ 3.2 degrees (approximately)
Therefore, the angle between the wire and the magnetic field is approximately 3.2 degrees.
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The angle is approximately 6.6°.
The formula for finding the magnetic force acting on a current carrying conductor in a magnetic field is,
F = BILSinθ Where,
F is the magnetic force in Newtons,
B is the magnetic field in Tesla
I is the current in Amperes
L is the length of the conductor in meters and
θ is the angle between the direction of current flow and the magnetic field lines.
Substituting the given values, we have,
F = 0.55 NB
= 1.25 TI
= 6.5 AL
= 45/100 meters (0.45 m)
Let θ be the angle between the wire and the 1.25 T field.
The force equation becomes,
F = BILsinθ 0.55
= (1.25) (6.5) (0.45) sinθ
sinθ = 0.55 / (1.25 x 6.5 x 0.45)
= 0.11465781711
sinθ = 0.1147
θ = sin^-1(0.1147)
θ = 6.6099°
= 6.6°
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An ion source is producing "Li ions, which have charge +e and mass 9.99 x 10-27 kg. The ions are accelerated by a potential difference of 15 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.0 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the "Li ions to pass through undeflected. = Number Units
The electric field necessary to counteract the magnetic force is calculated using the formula F = EqR, where F is the force, E is the electric field strength, and R is the radius of the circular path the ions would follow without the electric field.
The strength of the smallest electric field required to allow Li+ ions to pass through the region without deflection is determined by balancing the magnetic force and the electric force.
Given that the radius of the circular path should be infinite for the ions to pass undeflected, the force required is zero. However, due to the need for the ions to be accelerated, a small electric field must be present.
Using the equation E = F/R and substituting the given values, we find that E = (2qV/m) / 1000, where q is the charge of the ions, V is the potential difference, and m is the mass of the ions.
By substituting the known values, E = (2 × 1.60 × 10^-19 C × 15000 V / 9.99 × 10^-27 kg) / 1000 = 0.048 V/m = 48 mV/m.
Therefore, the smallest electric field strength required for the Li+ ions to pass through the region undeflected is 48 mV/m or 0.048 V/m.
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Suppose the yellow clip in the above image is attached to the G+ input on your iOLab, and the black clip is attached to the G-input, and that the High Gain sensor was being recorded during the flip. Describe what you think the High Gain data chart looks like. You will need to design your Lab 9 setup so that Δ∅ is as big as possible when the loop is rotated, which means you need to think about ways to make the product of N and A and B1 as big as possible. Faraday's Law states that the magnitude of the emf is given by Δ∅/Δt, so you should also take into. account the time it takes you to flip the loop. Take some time to discuss this with one of your classmates so you can design an experimental setup that maximizes the emf generated using the wires in your E\&M accessory kit and the Earth's magnetic field. 4. In the space below, summarize your thoughts and reasoning from your discussion with your classmate. Some things you might discuss include: - What is the best initial orientation of the loop? - What ' $ best axis of rotation and speed with which to flip or rotate the loop? - Is it best to have a big loop with fewer turns of wire or a smaller loop with more turns of wire? (Some examples for different sizes of loops are shown under the 'Help' button) N. Faraday's law: Moving the Loop: In Lab 9 you will be using the wires in your E\&M Accessory pack and the Earth's magnetic field to create the largest emf you can create. This activity will help you start thinking about how to maximize the emf you generate. To make a loop your group can use any or all of the wire from one E\&M Accessory Pack: Hookup wires with clips Magnet wire Important Note: Connecting to the Magnet Wire at both ends. You will be using the Earth itself as the magnet. Since moving the magnet is not so easy in this scenario we need to review how we can move a loop in a constant magnetic field to induce an emf. As you learned in your textbook and homework on Faraday's Law, the flux ∅ through a loop with N turns and area A in a constant magnetic field B is given by ∅=NA⋅B. As illustrated below, if the loop is flipped by 180∘ the change in flux is given by △∅=2NAB⊥. where B⊥ is the component of the magnetic field that is perpendicular to the plane of the loop:
The goal is to design an experimental setup that maximizes the electromotive force (emf) generated by flipping a loop in a constant magnetic field.
Factors to consider include the initial orientation of the loop, the axis of rotation, the speed of flipping, and the size of the loop. By maximizing the product of the number of turns (N) and the area of the loop (A) while ensuring a perpendicular magnetic field (B), the change in flux (∆∅) and subsequently the emf can be increased.
To maximize the emf generated, several considerations need to be made. Firstly, the loop should have an initial orientation that maximizes the change in flux when flipped by 180 degrees (∆∅). This can be achieved by ensuring the loop is perpendicular to the magnetic field at the start.
Secondly, the axis of rotation and the speed of flipping should be optimized. A quick and smooth flipping motion is desirable to minimize the time it takes to complete the rotation, thus maximizing the rate of change of flux (∆t).
Lastly, the size of the loop should be considered. Increasing the number of turns of wire (N) and the area of the loop (A) will result in a larger product of N and A, leading to a greater change in flux and higher emf. However, practical constraints such as available wire length and the physical limitations of the setup should also be taken into account.
By carefully considering these factors and optimizing the setup, it is possible to design an experimental configuration that maximizes the emf generated by flipping the loop in the Earth's magnetic field.
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An 7.20 kg package in a mail-sorting room slides 2.10 m down a chute that is inclined at 53.8 degrees below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.36. Calculate the work done on the package by
a) friction.
b) gravity.
c) the normal force
d) what is the net work done on the package?
The work done on the package by:a) friction: -228.024 J b) gravity: -348.634 Jc) the normal force: 0 J d) the net work done on the package: -576.658 J
a) The work done by friction can be calculated using the equation W_friction = -μk * N * d, where μk is the coefficient of kinetic friction, N is the normal force, and d is the displacement. The negative sign indicates that the work done by friction is in the opposite direction of the displacement.
b) The work done by gravity can be calculated using the equation W_gravity = m * g * d * cos(θ), where m is the mass of the package, g is the acceleration due to gravity, d is the displacement, and θ is the angle of the incline. The cos(θ) term accounts for the component of gravity parallel to the displacement.
c) The work done by the normal force is zero because the displacement is perpendicular to the direction of the normal force.
d) The net work done on the package is the sum of the work done by friction and the work done by gravity, i.e., W_net = W_friction + W_gravity. It represents the total energy transferred to or from the package during its motion along the chute.
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A generating station is producing 1.1×106 W of power that is to be sent to a small town located 6.8 km away. Each of the two wires that comprise the transmission line has a resistance per length of 5.0×10−2 d/km. (a) Find the power lost in heating the wires if the power is transmitted at 1600 V. (b) A 100:1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now lost in heating the wires? (a) Number Units (b) Number Units
(a) 150W
(b) 31858.20 W (approximately)
(a) Let's find the power loss in heating the wires if the power is transmitted at 1600 V.
As we know that P = I²R ,
Where,
P = Power,
I = Current,
R = Resistance
As we know that,
V = IR ,
where,
V = Voltage,
I = Current,
R = Resistance
R = ρ l/A ,
where,
ρ = Resistivity,
l = Length,
A = Area
Therefore, P = I²ρ l/A or P = V²/R ,
where,
V = Voltage,
R = Resistance
P = (1600)²/(2 x 5.0×10−2 d x 6.8 km) = 150 W
(b) Now, let's find the power loss in heating the wires if 100:
1 step-up transformer is used to raise the voltage before the power is transmitted.
Therefore, the new voltage, V = 1600 x 100
= 160000V, and
the new current, I = 1.1×10⁶ / 160000
= 6.875A.
Now,
resistance,
R = 2 x 5.0×10−2 d x 6.8 km
= 680 Ohms
P = I²R
= (6.875)² x 680 = 31858.20 W
Therefore, the power loss in heating the wires after using the transformer is 31858.20 W (approximately).
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A hockey player with a mass of 62 kg is skating with an initial velocity of 5.7 m/s [N26°E] when she collides with another hockey player with a mass of 53 kg travelling with a velocity of 3.8 m/s [N]. If the heavier hockey player has a velocity of 5.0 m/s [N11°E] after the collision, determine the final velocity of the 53 kg hockey player. 4.4 m/s [N24°E] 2.0 m/s [S18°E] 5.4 m/s [N23°E] 3.1 m/s [S7.2°E]
The final velocity of the 53 kg hockey player after the collision is approximately 3.1 m/s [S7.2°E]. We can apply the principle of conservation of momentum.
Momentum is defined as the product of mass and velocity, and the total momentum before the collision should be equal to the total momentum after the collision.
Let's break down the initial velocities of both players into their horizontal and vertical components.
The 62 kg player has an initial horizontal velocity = 5.7 m/s × cos(26°) and a vertical velocity = 5.7 m/s × sin(26°).
The 53 kg player has an initial horizontal velocity of 3.8 m/s and no vertical velocity.
Using the conservation of momentum, we can write the equation:
(mass of 62 kg player × horizontal velocity of 62 kg player) + (mass of 53 kg player × horizontal velocity of 53 kg player) = (mass of 62 kg player × final horizontal velocity of 62 kg player) + (mass of 53 kg player × final horizontal velocity of 53 kg player)
(62 kg × 5.7 m/s × cos(26°)) + (53 kg × 3.8 m/s) = (62 kg × final horizontal velocity of 62 kg player × cos(11°)) + (53 kg × final horizontal velocity of 53 kg player)
Simplifying the equation, we can solve for the final horizontal velocity of the 53 kg player:
(62 kg × 5.7 m/s × cos(26°)) + (53 kg × 3.8 m/s) = (62 kg × 5.0 m/s × cos(11°)) + (53 kg × final horizontal velocity of 53 kg player
After calculating the values on the left side and rearranging the equation, we find that the final horizontal velocity of the 53 kg player is approximately 3.1 m/s.
To determine the direction, we use trigonometry to find the angle:
final angle = a tan((53 kg × final horizontal velocity of 53 kg player) / (62 kg × 5.0 m/s × sin(11°)))
Calculating the value, we get an angle of approximately 7.2° south of the positive x-axis.
Therefore, the final velocity of the 53 kg hockey player is approximately 3.1 m/s [S7.2°E].
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we have a rod-shaped space station of length 714 m and mass 9.69 x 10^6 kg, which can change its length (kind of like an old-fashioned telescope), without changing its overall mass. Suppose that the station is initially rotating at a constant rate of 1.32 rpm. If the length of the rod is reduced to 1.32 m, what will be the new rotation rate of the space station?
A. 6.21 rpm
B. 2.03 rpm
C. 4.14 rpm
D. 2.90 rpm
The option B is correct. The new rotation rate of the space station is approximately 2.03 rpm.
Let I1 be the moment of inertia of the space station initially.I1 = (1/3) M L²When the length is reduced to 1.32 m, let I2 be the new moment of inertia.I2 = (1/3) M L'²where L' is the new length of the space station. The moment of inertia of the space station varies as the square of the length of the rod.I1/I2 = L²/L'²I1 = I2 (L/L')²9.69 x 10^6 x (714)² = I2 (1.32)²I2 = 9.69 x 10^6 x (714)² / (1.32)²I2 = 1.138 x 10^6 kg m².
The initial angular velocity of the space station, ω1 = 1.32 rpmω1 = (2π / 60) rad/sω1 = (π / 30) rad/s. The law of conservation of angular momentum states that the initial angular momentum of the space station is equal to the final angular momentum of the space station.I1 ω1 = I2 ω2(1/3) M L² (π / 30) = 1.138 x 10^6 ω2ω2 = (1/3) M L² (π / 30) / I2ω2 = (1/3) (9.69 x 10^6) (714)² (π / 30) / (1.138 x 10^6)ω2 = 2.03 rpm. Therefore, the new rotation rate of the space station is 2.03 rpm (approximately).
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A soccer ball that has just been kicked by Lionel Messi has a kinetic energy of 1440 J and has a mass of 450 g. What velocity is the soccer ball travelling at? O / A. 56 m/s O s B. 75 m/s O C./ 80 m/s OD. 12 m/
The soccer ball is traveling at approximately 53.67 m/s. Option A is correct.
To calculate the velocity of the soccer ball, we can use the formula for kinetic energy:
Kinetic energy (KE) = (1/2) × mass × velocity²
Kinetic energy (KE) = 1440 J
Mass (m) = 450 g
= 0.45 kg
Rearranging the equation and solving for velocity (v):
KE = (1/2) × m × v²
1440 J = (1/2) × 0.45 kg × v²
Dividing both sides of the equation by (1/2) × 0.45 kg:
2880 J/kg = v²
Taking the square root of both sides:
v = √(2880 J/kg)
v = 53.67 m/s
Hence, Option A is correct.
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Consider the skier on a slope that is 32.8 degrees above horizontal. Her mass including equipment is 58.7 kg. E (a) What is her acceleration if friction is negligible? E a== units m/s^2
The acceleration of a skier on a slope that is 32.8 degrees above the horizontal is 3.66 m/s^2, assuming that the friction is negligible.
Let's derive this solution step by step. During free fall, acceleration is due to gravity. The acceleration due to gravity is 9.8 m/s^2 in the absence of air resistance. A component of the weight vector is applied parallel to the slope, resulting in a downhill acceleration.
The skier's weight is mg, where m is the mass of the skier and equipment and g is the acceleration due to gravity, which we assume to be constant.
Calculate the force parallel to the slope, which is the force acting to propel the skier forward down the slope. The downhill force is equivalent to the force acting along the x-axis, which is directed parallel to the slope. When we resolve the weight into components perpendicular and parallel to the slope,
The parallel component is : Parallel Force = Weight × sin (32.8).
We assume that the friction force is negligible since we are told to disregard it in the problem statement. The downhill acceleration is then obtained by dividing the downhill force by the skier's mass. It's expressed in meters per second squared
.Downhill Acceleration = (Parallel Force) / Mass = Weight × sin (32.8) / Mass
= (58.7 kg × 9.8 m/s^2 × sin 32.8) / 58.7 kg
= 3.66 m/s^2.
Therefore, the skier's acceleration is 3.66 m/s^2.
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#1 Consider the following charge distribution in the x-y plane. The first charge 1 =+ is placed at the position 1=(,0). A second 2 =− is placed at position 2 =(−,0), and a third charge 3 = +3 is placed at position 3 =(0,−). At =(0,0), solve for: (a) the electric field; (b) the electric potential. Take =2 nm, =3 nm, and =.
#2 A thin rod of length ℓ with positive charge distributed uniformly throughout it is situated horizontally in the x-y plane. Take it to be oriented along the x-axis such that its left end is at position x=−ℓ/2, and its right end is at position x=ℓ/2. At position =(−ℓ/2,ℎ), solve for: (a) the electric field; (b) the electric potential.
#3 If a point charge with charge − =− is positioned at x=−, where on the x-axis could you put a point charge with charge + =+3 such that: (a) the electric field at x=0 is zero? (b) the electric potential at x=0 is zero?
Thank you and please solve all questions!
Question #1 involves a charge distribution in the x-y plane, where three charges are placed at specific positions. The task is to determine the electric field and electric potential at the origin (0,0). Question #2 deals with a thin rod of positive charge placed horizontally in the x-y plane, and the goal is to find the electric field and electric potential at a given position. In Question #3, a point charge with a negative charge is positioned at a specific point on the x-axis, and the objective is to determine where a point charge with a positive charge should be placed so that the electric field or electric potential at the origin (x=0) is zero.
For Question #1, to find the electric field at the origin, we need to consider the contributions from each charge and their distances. The electric field due to each charge is given by Coulomb's law, and the total electric field at the origin is the vector sum of the electric fields due to each charge. To find the electric potential at the origin, we can use the principle of superposition and sum up the electric potentials due to each charge.
In Question #2, to determine the electric field at a given position (x,h), we need to consider the contributions from different sections of the rod. We can divide the rod into small segments and calculate the electric field due to each segment using Coulomb's law. The total electric field at the given position is the vector sum of the electric fields due to each segment. To find the electric potential at the given position, we can integrate the electric field along the x-axis from the left end of the rod to the given position.
For Question #3(a), to have zero electric field at x=0, we need to place the positive charge at a point where the electric field due to the positive charge cancels out the electric field due to the negative charge. The distances between the charges and the position of the positive charge need to be taken into account. For Question #3(b), to have zero electric potential at x=0, we need to place the positive charge at a position where the electric potential due to the positive charge cancels out the electric potential due to the negative charge. Again, the distances between the charges and the position of the positive charge must be considered.
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If electrical energy costs $0.12 per kilowatt-hour, how much do the following events cost? (a) To burn a 80.0-W lightbulb for 24 h. (b) To operate an electric oven for 5.3 h if it carries a current of 20.0 A at 220 V.
(a) To burn a 80.0-W lightbulb for 24 h costs $0.96.
(b) To operate an electric oven for 5.3 h if it carries a current of 20.0 A at 220 V costs $1.24.
Here are the details:
The cost of burning a 80.0-W lightbulb for 24 h is calculated as follows:
Cost = Power * Time * Cost per kilowatt-hour
where:
* Cost is in dollars
* Power is in watts
* Time is in hours
* Cost per kilowatt-hour is in dollars per kilowatt-hour
In this case, the power is 80.0 W, the time is 24 h, and the cost per kilowatt-hour is $0.12. Plugging in these values, we get:
Cost = 80.0 W * 24 h * $0.12/kWh = $0.96
The cost of operating an electric oven for 5.3 h if it carries a current of 20.0 A at 220 V is calculated as follows:
Cost = Current * Voltage * Time * Cost per kilowatt-hour
where:
* Cost is in dollars
* Current is in amperes
* Voltage is in volts
* Time is in hours
* Cost per kilowatt-hour is in dollars per kilowatt-hour
In this case, the current is 20.0 A, the voltage is 220 V, the time is 5.3 h, and the cost per kilowatt-hour is $0.12. Plugging in these values, we get:
Cost = 20.0 A * 220 V * 5.3 h * $0.12/kWh = $1.24
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Question 12 If a proton is in an infinite box in the n = 4 state and its energy is 0.662 MeV, what is the wavelength of this proton (in fm)? 1 pts
The wavelength of the proton in the n = 4 state in the infinite box is approximately 19.55 femtometers (fm). To determine the wavelength of a proton in an infinite box in the n = 4 state, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength.
The de Broglie wavelength (λ) is given by:
λ = h / p
where λ is the wavelength, h is the Planck's constant (approximately 6.626 × 10^-34 J·s), and p is the momentum of the proton.
The momentum of the proton can be calculated using the energy (E) and mass (m) of the proton:
E = [tex]p^2 / (2m)[/tex]
Rearranging the equation, we can solve for p:
p = √(2mE)
Energy of the proton (E) = 0.662 MeV = 0.662 × [tex]10^6[/tex] eV
Mass of the proton (m) = 1.67 × [tex]10^-27[/tex] kg
Converting the energy to joules:
1 eV = 1.6 × [tex]10^-19[/tex] J
E = 0.662 ×[tex]10^6[/tex] eV * (1.6 × [tex]10^-19[/tex] J / 1 eV)
E = 1.0592 ×[tex]10^-13[/tex] J
Now we can calculate the momentum:
p = √(2mE)
p = √(2 * 1.67 × [tex]10^-27[/tex] kg * 1.0592 × [tex]10^-13[/tex]J)
p ≈ 3.382 × [tex]10^-20[/tex] kg·m/s
Finally, we can calculate the wavelength using the de Broglie wavelength formula:
λ = h / p
λ = (6.626 × [tex]10^-34[/tex] J·s) / (3.382 × [tex]10^-20[/tex]kg·m/s)
λ ≈ 1.955 × 10^-14 m
Converting the wavelength to femtometers (fm):
1 m = [tex]10^15[/tex] fm
λ = 1.955 × [tex]10^-14[/tex]m * [tex](10^{15[/tex] fm / 1 m)
λ ≈ 19.55 fm
Therefore, the wavelength of the proton in the n = 4 state in the infinite box is approximately 19.55 femtometers (fm).
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How much heat must be added to 7kg of water at a temperature of
18°C to convert it to steam at 133°C
The amount of heat required to convert 7kg of water at a temperature of 18°C to convert it to steam at 133°C is 18713.24 kJ.
To calculate the amount of heat required to convert water at a certain temperature to steam at another temperature, we need to consider two steps:
heating the water from 18°C to its boiling point and then converting it to steam at 100°C, and
then heating the steam from 100°C to 133°C.
Heating water to boiling pointThe specific heat capacity of water is approximately 4.18 J/g°C.
The boiling point of water is 100°C, so the temperature difference is 100°C - 18°C = 82°C.
The heat required to raise the temperature of 7 kg of water by 82°C can be calculated using the formula:
Heat = mass * specific heat capacity * temperature difference
Heat = 7 kg * 4.18 J/g°C * 82°C = 2891.24 kJ
Converting water to steamTo convert water to steam at its boiling point, we need to consider the heat of the vaporization of water. The heat of the vaporization of water is approximately 2260 kJ/kg.
The heat required to convert 7 kg of water to steam at 100°C can be calculated using the formula:
Heat = mass * heat of vaporization
Heat = 7 kg * 2260 kJ/kg = 15820 kJ
Heating steam from 100°C to 133°CThe specific heat capacity of steam is approximately 2.0 J/g°C.
The temperature difference is 133°C - 100°C = 33°C.
The heat required to raise the temperature of 7 kg of steam by 33°C can be calculated using the formula:
Heat = mass * specific heat capacity * temperature difference
Heat = 7 kg * 2.0 J/g°C * 33°C = 462 J
Total heat required = Heat in Step 1 + Heat in Step 2 + Heat in Step 3
Total heat required = 2891.24 kJ + 15820 kJ + 462 J = 18713.24 kJ
Therefore, approximately 18713.24 kJ of heat must be added to convert 7 kg of water at a temperature of 18°C to steam at 133°C.
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Around the star Kepler-90, a system of planets has been detected.
The outermost two (Kepler-90g & Kepler-90h) lie at an average of 106 Gm and and 151 Gm from the central star, respectively.
From the vantage point of the exoplanet Kepler-90g, an orbiting moon around Kepler-90h will have a delay in its transits in front of Kepler-90h due to the finite speed of light.
The speed of light is 0.300 Gm/s. What will be the average time delay of these transits in seconds when the two planets are at their closest?
The average time delay of the transits of Kepler-90h from the perspective of Kepler-90g, caused by the finite speed of light, will be approximately 857.33 seconds when the two planets are at their closest.
To calculate the average time delay of the transits of Kepler-90h caused by the finite speed of light from the perspective of Kepler-90g, we need to determine the time it takes for light to travel the distance between the two planets when they are at their closest.
Given:
Distance between Kepler-90g and Kepler-90h at their closest (d) = 106 Gm + 151 Gm = 257 Gm
Speed of light (c) = 0.300 Gm/s
Time delay (Δt) can be calculated using the formula:
Δt = d / c
Substituting the given values:
Δt = 257 Gm / 0.300 Gm/s
Δt = 857.33 s
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What is the angular velocity of a tricycle wheel relative to the angular velocity of a bicycle wheel (what is w tricycle/w bicycle) if both wheels are traveling with the same translational speed? The bicycle has a wheel radius that is 3.00 times that of the tricycle wheel. Would it be safe to make a child tricycle/adult bicycle tandem?
The angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3) and it would not be safe to make a child tricycle/adult bicycle tandem.
To determine the angular velocity ratio between the tricycle wheel and the bicycle wheel, we can use the relationship between linear speed, angular velocity, and the radius of a rotating object.
The linear speed of both wheels is the same since they are traveling at the same translational speed.
Let's denote the linear speed as v.
For the bicycle wheel, let's denote its radius as r_bicycle.
For the tricycle wheel, let's denote its radius as r_tricycle.
The relationship between linear speed and angular velocity is given by:
v = ω * r,
where v is the linear speed, ω (omega) is the angular velocity, and r is the radius of the rotating object.
For the bicycle wheel, we have:
v_bicycle = ω_bicycle * r_bicycle.
For the tricycle wheel, we have:
v_tricycle = ω_tricycle * r_tricycle.
Since both wheels have the same linear speed, we can set the two equations equal to each other:
v_bicycle = v_tricycle.
ω_bicycle * r_bicycle = ω_tricycle * r_tricycle.
We can rewrite this equation in terms of the angular velocity ratio:
ω_tricycle / ω_bicycle = r_bicycle / r_tricycle.
Given that the radius of the bicycle wheel is 3.00 times that of the tricycle wheel (r_bicycle = 3 * r_tricycle), we can substitute this into the equation:
ω_tricycle / ω_bicycle = (3 * r_tricycle) / r_tricycle.
ω_tricycle / ω_bicycle = 3.
Therefore, the angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3).
Based on this, it would not be safe to make a child tricycle/adult bicycle tandem because the tricycle wheel would rotate at a much higher angular velocity than the bicycle wheel, potentially causing stability issues and safety concerns.
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A police officer is driving his car with a speed of 20 mph; he is using a radar in X band with a frequency of 10 GHz to determine the speeds of moving vehicles behind
him. If the Doppler shift on his radar is 2.00 KHz. Find the speed in mph
(a) for a vehicle moving in the same direction? (b) for a vehicle moving in the opposite direction?
Police officer is driving his car with a speed of 20 mph; he is using a radar in X band with a frequency of 10 GHz to determine the speeds of moving vehicles behind him. If the Doppler shift on his radar is 2.00 KHz.(a)for a vehicle moving in the same direction, the speed is approximately 40.32 mph.(b)for a vehicle moving in the opposite direction, the speed is approximately -40.32 mph. The negative sign indicates the opposite direction of motion
(a) For a vehicle moving in the same direction:
Given:
Speed of the police officer's car = 20 mph
Frequency shift observed (Δf) = 2.00 KHz = 2.00 x 10^3 Hz
Original frequency emitted by the radar (f₀) = 10 GHz = 10^10 Hz
To calculate the speed of the vehicle in the same direction, we can use the formula:
Δf/f₀ = v/c
Rearranging the equation to solve for the speed (v):
v = (Δf/f₀) × c
Substituting the values:
v = (2.00 x 10^3 Hz) / (10^10 Hz) × (3.00 x 10^8 m/s)
Converting the speed to miles per hour (mph):
v = [(2.00 x 10^3 Hz) / (10^10 Hz)× (3.00 x 10^8 m/s)] × (2.24 mph/m/s)
Calculating the speed:
v ≈ 40.32 mph
Therefore, for a vehicle moving in the same direction, the speed is approximately 40.32 mph.
(b) For a vehicle moving in the opposite direction:
Given the same values as in part (a), but now we need to consider the opposite direction.
Using the same formula as above:
v = (Δf/f₀) × c
Substituting the values:
v = (-2.00 x 10^3 Hz) / (10^10 Hz) × (3.00 x 10^8 m/s)
Converting the speed to miles per hour (mph):
v = [(-2.00 x 10^3 Hz) / (10^10 Hz) × (3.00 x 10^8 m/s)] × (2.24 mph/m/s)
Calculating the speed:
v ≈ -40.32 mph
Therefore, for a vehicle moving in the opposite direction, the speed is approximately -40.32 mph. The negative sign indicates the opposite direction of motion.
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A rectangular piece of wood floats in water of density 1000kg/m^3 . Bath oil of density 300kg/m^3 is slowly added, forming a layer that does not mix with the water. When the top surface of the oil is at the same level as the top surface of the wood, the ratio of the oil layer thickness to the wood’s thickness, x/L = 4/7 . What is the density of the wood?
Answer:
Mass = density * volume = ρ V
Mass of wood = Mass Water + Mass Oil (multiply by g to get weight)
Vw ρw = 3/7 V (1000 kg/m^3) + 4/7 V 300 kg/m^3)
Let V be 1
ρw = (3000 + 1200) kg/m^3/ 7 = 600 kg/m^3
Density = 600 kg/m^3
Tutorial 2 (Centrifugal Pump) A centrifugal pump with outlet diameter of 400 mm and the width of outlet impeller 15 mm is required to produce manometric head of H = 60+ 500Q². The inlet diameter of the pump is 200 mm can be operated with N=1450 rpm with the backward-curved impeller of B₂=45°. The impeller blades occupy 10% of the circumference. The manometric and overall efficiencies of the pump are 85% and 75%, respectively. Determine: a. Q b. Power input c. Blade angle at the inlet.
a. The flow rate (Q) can be determined by rearranging the
given equation for manometric.
Rearranging the equation gives:
500Q² = H - 60
Q² = (H - 60) / 500
Taking the square root of both sides:
Q = √((H - 60) / 500)
Substituting the given value of H (60 + 500Q²) into the equation will provide the flow rate (Q).
b. The power input to the pump can be calculated using the following formula:
P = (ρQH) / (ηmηo)
Where:
P = Power input to the pump
ρ = Density of the fluid
Q = Flow rate
H = Manometric head
ηm = Manometric efficiency
ηo = Overall efficiency
Substituting the given values into the formula will yield the power input (P) in the appropriate units.
c. The blade angle at the inlet can be determined by using the backward-curved impeller configuration and the percentage of blade occupancy. In a backward-curved impeller, the blades curve away from the direction of rotation. The blade angle at the inlet is given by:
β₁ = β₂ - (180° / π) * (2θ / 360°)
Where:
β₁ = Blade angle at the inlet
β₂ = Blade angle at the outlet
θ = Percentage of blade occupancy (given as 10%)
By substituting the given values into the equation, the blade angle at the inlet (β₁) can be calculated.
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1.Find the force on a particle of mass m=1.70×10-27kg and charge q=1.60×10-19C if it enters a field B=5 mT with an initial speed of v=83.5 km/s. Assume the velocity is in the x direction and the magnetic field enters perpendicular to the screen. Also make a schematic drawing of these vectors. Don't forget to place your reference system.
2.Find the force on a straight conductor of length 0.3 m, which carries a current of 5 A in the negative z-direction. In that space there is a magnetic field given by the vector B=3.5×10-3Ti-3.5×10-3Tj . Make a schematic drawing of the situation. (We do not require precision in your drawing for the direction of the magnetic field, only approximate).
3.A conductor of length 2.5 m is located at z=0, x=4m with a current of 12 A in the -y direction. Find the magnetic field that exists in that region if the force on the conductor is F=-1.20×10-2N(-12i-12j).
4.A long thin wire carries a current I. A metal bar of length L is moving with a constant speed v as shown in the figure. Point a is a distance d from the wire a) Calculate the electromotive force induced in the bar. b) If the bar is replaced by a rectangular circuit of resistance R, what is the magnitude of the induced current in the circuit?
1. The force on the particle is 1.36 x 10^-14 N, schematic drawing shows velocity in x-direction, magnetic field entering perpendicular to the screen, and force perpendicular to both.
2. The force on the straight conductor is 5.25 x 10^-3 N, schematic drawing shows conductor in negative z-direction and magnetic field vectors approximately orthogonal to the conductor.
3. The magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.
4. a) The electromotive force induced in the bar is BLv. b) The magnitude of the induced current in the rectangular circuit is V/R.
1. The force on the particle can be calculated using the equation F = qvB, where q is the charge, v is the velocity, and B is the magnetic field. Plugging in the given values, the force is 1.36 x 10^-14 N. A schematic drawing would show the velocity vector in the x-direction, the magnetic field vector entering perpendicular to the screen, and the force vector perpendicular to both.
2. The force on the straight conductor can be calculated using the equation F = IL x B, where I is the current, L is the length of the conductor, and B is the magnetic field. Plugging in the given values, the force is 5.25 x 10^-3 N. A schematic drawing would show the conductor in the negative z-direction, with the magnetic field vectors shown approximately orthogonal to the conductor.
3. The magnetic field can be determined using the equation F = IL x B. Since the force is given as F = -1.20 x 10^-2 N (-12i - 12j), we can equate the force components to the corresponding components of the equation and solve for B. The resulting magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.
4. To calculate the electromotive force induced in the bar, we can use the equation emf = BLv, where B is the magnetic field, L is the length of the bar, and v is the speed of the bar. The magnitude of the induced current in the rectangular circuit can be calculated using Ohm's Law, I = V/R, where V is the electromotive force and R is the resistance of the circuit.
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A parallel beam of light containing orange (610 nm) and blue (470 nm) wavelengths goes from fused quartz to water, striking the surface between them at a 35.0° incident angle. What is the angle between the two colors in water? Submit Answer Incorrect. Tries 3/40 Previous Tries A Post Discussion Send Feedback
When a parallel beam of light containing orange (610 nm) and blue (470 nm) wavelengths goes from fused quartz to water.
striking the surface between them at a 35.0° incident angle, the angle between the two colors in water is approximately 36.8°.Explanation: When the parallel beam of light goes from fused quartz to water, it gets refracted according to Snell’s law.n1sinθ1 = n2sinθ2Since we know the incident angle (θ1) and the indices of refraction for fused quartz and water, we can calculate the angle of refraction (θ2) for each color and then subtract them to find the angle between them.θ1 = 35.0°n1 (fused quartz) = 1.46n2 (water) = 1.33.
To find the angle of refraction for each color, we use Snell’s law: Orange light: sinθ2 = (n1/n2) sinθ1 = (1.46/1.33) sin(35.0°) = 0.444θ2 = sin−1(0.444) = 26.1°Blue light: sinθ2 = (1.46/1.33) sin(35.0°) = 0.532θ2 = sin−1(0.532) = 32.5°Therefore, the angle between the two colors in water is:32.5° − 26.1° ≈ 6.4° ≈ 36.8° (to one decimal place)Answer: Approximately 36.8°.
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Q/C A man claims that he can hold onto a 12.0 -kg child in a head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 60.0m/h relative to the ground. The car in which he rides is brought to rest in 0.10s . (c) What does the answer to this problem say about laws requiring the use of proper safety devices such as seat belts and special toddler seats?
The man's claim about holding onto a 12.0-kg child in a head-on collision is incorrect. In this scenario, both cars are traveling at 60.0 m/h relative to the ground and collide. The car the man is in is brought to rest in 0.10s.
To assess the situation, we can use the principle of conservation of momentum. The total momentum of the system before the collision should be equal to the total momentum after the collision.
Since the cars are identical, they have the same mass and are traveling at the same speed in opposite directions. Therefore, the total momentum before the collision is zero.
After the collision, the car the man is in comes to a stop, resulting in a change in momentum. This means that the total momentum after the collision is not zero.
The fact that the car comes to a stop in such a short time demonstrates the significant forces involved in the collision. If the man were holding onto the child without a seat belt, both of them would experience an abrupt change in momentum and could be seriously injured or thrown from the car.
This problem emphasizes the importance of laws requiring the use of proper safety devices such as seat belts and special toddler seats. These devices help to distribute the forces of a collision more evenly throughout the body, reducing the risk of injury.
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A cannonball is falling from rest; air resistance is
considered. Before the cannonball reaches terminal velocity, the
cannonball is speeding up. Explain why.
The cannonball is speeding up before it reaches terminal velocity due to the presence of gravitational force.
When the cannonball is initially dropped, gravity pulls it downward, and it begins to accelerate. At this stage, the air resistance opposing the motion is relatively low because the speed of the falling cannonball is still relatively low. As the cannonball accelerates, its speed increases, and the air resistance acting against it also increases. Air resistance is a force that opposes the motion of an object through the air, and it depends on factors such as the shape, size, and speed of the object. Initially, the air resistance is not strong enough to counteract the gravitational force pulling the cannonball downward. However, as the cannonball gains speed, the air resistance becomes stronger. Eventually, the air resistance force becomes equal to the gravitational force, and the cannonball reaches its terminal velocity. At this point, the forces acting on the cannonball are balanced, resulting in a constant velocity. Therefore, until the cannonball reaches its terminal velocity, the gravitational force is greater than the opposing air resistance, causing the cannonball to accelerate and speed up.
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Aray of light impinges on a mirror as shown in (Eigure 1) = 40" A second mirror is fastened at 90 to the first Part A e Figure s not At what angle above horizontal does the ray emerge after reflecting from both mirrors? Express your answer in degrees. VAX ? 0 - 170 Submit Previous Answers Request Answer
The angle above horizontal at which the ray emerges after reflecting from both mirrors is 50 degrees.
When a ray of light impinges on the first mirror at an angle of 40 degrees, it reflects at the same angle due to the law of reflection. Now, the second mirror is fastened at a 90-degree angle to the first mirror, which means the ray will reflect vertically upwards.
To find the angle above horizontal at which the ray emerges, we need to consider the angle of incidence on the second mirror. Since the ray is reflected vertically upwards, the angle of incidence on the second mirror is 90 degrees.
Using the principle of alternate angles, we can determine that the angle of reflection on the second mirror is also 90 degrees. Now, the ray is traveling in a vertical direction.
To find the angle above horizontal, we need to measure the angle between the vertical direction and the horizontal direction. Since the vertical direction is perpendicular to the horizontal direction, the angle above horizontal is 90 degrees.
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