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Crank
4 years ago
12

A boxcar at a rail yard is set into motion at the top of a hump. The car rolls down quietly and without friction onto a straight

, level track where it couples with a flatcar of smaller mass, originally at rest, so that the two cars then roll together without friction. Consider the two cars as a system from the moment of release of the boxcar until both are rolling together. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved? Next, consider only the process of the boxcar gaining speed as it rolls down the hump. Consider the boxcar and the Earth as a system. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved? Finally, consider the two cars as a system as the boxcar is slowing down in the coupling process. (a) is the mechanical energy of the system conserved? (b) is the momentum of this system conserved?

Physics
1 answer:
enyata [817]4 years ago
3 0

Answer:

I have answer with explanations in each of the six (6) special cases.

Explanation:

Please find attached the detailed solution to all your questions.

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Am radio signals have frequencies between 550 khz and 1600 khz (kilohertz) and travel with a speed of 3.00 ✕ 108 m/s. what are t
Allisa [31]
The minimum frequency is 
f_1 = 550 kHz = 5.50 \cdot 10^5 Hz
while the maximum frequency is
f_2 = 1600 kHz = 1.6 \cdot 10^6 Hz
Using the relationship between frequency f of a wave, wavelength \lambda and the speed of the wave v, we can find what wavelength these frequencies correspond to:
\lambda_1 =  \frac{v}{f_1}= \frac{3 \cdot 10^8 m/s}{5.5 \cdot 10^5 Hz}=545 m
\lambda_2 =  \frac{v}{f_2}= \frac{3 \cdot 10^8 m/s}{1.6 \cdot 10^6 Hz}=188 m

So, the wavelengths of the radio waves of the problem are within the range 188-545 m.

5 0
3 years ago
Three 500-g point masses are at the corners of an equilateral triangle with 50-cm sides. What is the moment of inertia of this s
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Answer:

0.25 kg m^2

Explanation:

mass of each , m = 500 g = 0.5 kg

distance, r = 50 cm = 0.5 m

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I = mr^2 + mr^2

I = 2 mr^2

I = 2 x 0.5 x 0.5 x 0.5

I = 0.25 kgm^2

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D, small force for short time, as it has the least influence on the velocity. 
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spayn [35]

Explanation:

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