The average acceleration during skiing is 0.4 [tex]m/s^2.[/tex]
To calculate the average acceleration, we can use the following equation:
average acceleration = (final velocity - initial velocity) / time
We can assume that the initial velocity is 0 m/s since we start from rest. We need to find the time it takes to travel the horizontal distance of 500 m.
To do this, we can use the following equation:
distance = average velocity x time
We can calculate the average velocity as:
[tex]average velocity = (0 m/s + 20 m/s) / 2 \\= 10 m/s[/tex]
Substituting this and the distance of 500 m into the equation above, we get:
[tex]500 m = 10 m/s * time[/tex]
Solving for time, we get:
[tex]time = 500 m / 10 m/s \\= 50 s[/tex]
Now we can calculate the average acceleration as:
average acceleration = (final velocity - initial velocity) / time
[tex]= (20 m/s - 0 m/s) / 50 s \\= 0.4 m/s^2[/tex]
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--The complete Question is, If you go skiing on a 1200 m vertical mountain and reach a final velocity of 20 m/s after traveling a horizontal distance of 500 m, what is your average acceleration? --
4. A 50 mH inductor is placed in parallel with a 100.0Ω resistor. Ignoring any resistance in the inductor itself, what is the time constant in seconds of this RL circuit
5.0 ms (milliseconds) is the time constant in seconds of this RL circuit.
The time constant of an RL circuit is given by the product of the resistance and the inductance, or τ = L/R. In this case, the inductance is 50 mH (millihenries), or 0.050 H, and the resistance is 100.0 Ω (ohms).
Plugging these values into the equation, we get:
[tex]τ = L/R = (0.050 H)/(100.0 Ω) = 0.0005 s = 0.5 ms[/tex]
Therefore, the time constant of the RL circuit is 0.5 ms (milliseconds), or 5.0 × 10^-4 seconds. This represents the time it takes for the current in the circuit to reach approximately 63% of its maximum value, or for the voltage across the inductor to reach approximately 63% of its maximum value when a DC voltage is initially applied to the circuit. The time constant is an important parameter in analyzing the transient behavior of an RL circuit.
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What is the frequency of light in a vacuum that has a wavelength of 70600 m?
(Choose from the following units: m, hz, m/s, s, degrees, dB)
Your Answer:
The frequency of light in a vacuum that has a wavelength of 70600 m is approximately 4.25 × 10⁻⁶ Hz.
The relationship between the frequency (f), wavelength (λ), and the speed of light (c) is given by the equation:
c = fλwhere c is approximately equal to 3 × 10⁸ meters per second in a vacuum.
Rearranging this equation, we can solve for the frequency:
f = c / λPlugging in the given wavelength of 70600 m, we get:
f = (3 × 10⁸ m/s) / (70600 m) ≈ 4.25 × 10⁻⁶ HzTherefore, the frequency of light in a vacuum that has a wavelength of 70600 m is approximately 4.25 × 10⁻⁶ Hz.
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5. Find the resonance frequency Æ0 in Hz of a series LRC circuit with L=0.030H, C=1.0*10^-9, and R=1000 Ω.
The resonance frequency Æ0 is 318309.89 Hz. The resonance frequency of a series LRC circuit is given by the formula [tex]Æ0 = 1/(2π√(LC))[/tex].
Substituting the given values, we get [tex]Æ0 = 1/(2π√(0.030H x 1.0*10^-9F))[/tex] = [tex]318309.89 Hz[/tex]. This means that when an AC voltage is applied to the circuit at this frequency, the circuit will resonate and the current will be maximum. At frequencies higher or lower than the resonance frequency, the current will decrease. The resistor R in the circuit causes the current to decrease with time and thus limits the amplitude of the resonant current.
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Given that the planet orbiting the nearby star 51 Pegasi is about 20X larger than the Earth, but 400X more massive, on that world you would weigh: A. twice as much as you do here. B. 20X more that you do here. C. half as much as you do here. D. 400X more than you do here. E. the same as you do here.
The weight of me will be the same as I do here. So the correct option is E.
Your weight on a planet is determined by the gravitational force exerted on you, which depends on the planet's mass and its radius. In this case, the planet orbiting 51 Pegasi is 20 times larger (radius) and 400 times more massive than Earth. To calculate your weight on this planet, we'll use the formula:
Weight_on_Planet = (Weight_on_Earth × Mass_of_Planet) / (Radius_of_Planet^2)
Let's substitute the given values (20 times larger and 400 times more massive):
Weight_on_Planet = (Weight_on_Earth × 400) / ([tex]20^{2}[/tex])
Weight_on_Planet = (Weight_on_Earth × 400) / 400
Weight_on_Planet = Weight_on_Earth
So, on the planet orbiting 51 Pegasi, you would weigh the same as you do on Earth. Therefore, the answer is E. the same as you do here.
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42. A car is moving along a horizontal road at a constant velocity that is directed 45° south of east. What is the direction of the angular velocity of the wheels of the car?
A) 45° south of west
B) 45° north of west
C) 45° south of east
D) 45° north of east
E) due east
The direction of the angular velocity of the wheels of the car is the same as the direction of the velocity of the car, which is 45° south of east. Therefore, the answer is C) 45° south of east.
The direction of the angular velocity of the wheels of the car is perpendicular to the plane of rotation. In this case, since the car is moving along a horizontal road, the wheels are rotating in a horizontal plane. Therefore, the angular velocity will have a direction perpendicular to this horizontal plane, which is vertically upwards or downwards. However, the given options do not include vertical directions, so we can assume that the question intends to ask for the direction of the linear velocity of a point on the rim of the wheel (the tangential velocity), which is parallel to the horizontal plane.
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