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alex41 [277]
2 years ago
14

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe

ed of 0.850 m/s and air track cart B has a mass of 300 g and moves toward the left with a speed of 1.12 m/s. What is the total momentum of the system?
A) 0.719 kg•m/s toward the right
B) 0.719 kg•m/s toward the left
C) 0.750 kg•m/s toward the right
D) 0.047 kg•m/s toward the right
E) 0.750 kg•m/s toward the left
Physics
1 answer:
Blizzard [7]2 years ago
6 0

Answer:

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

Explanation:

Hi there!

The total momentum of the system is given by the sum of the momentum vectors of each cart. The momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

Then, the momentum of the system will be the momentum of cart A plus the momentum of cart B (let´s consider the right as the positive direction):

mA · vA + mB · Vb

0.450 kg · 0.850 m/s + 0.300 kg · (- 1.12 m/s) = 0.047 kg · m/s

The right answer is D) the total momentum of the system is 0.047 kg · m/s toward the right.

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Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

T^{2}=\frac{4\pi^{2}}{GM}r^{3}    (1)

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So for satellite X, the orbital period T_{X} is:

T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}    (2)

Where r_{X}=1.2(10)^{6}m

T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}    (3)

T_{X}=413.712 s    (4)

For satellite Y, the orbital period T_{Y} is:

T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}    (5)

Where r_{Y}=1.9(10)^{5}m

T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}    (6)

T_{Y}=26.064 s    (7)

This means T_{X}>T_{Y}

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<u>For Satellite X:</u>

V_{X}=\sqrt{\frac{GM}{r_{X}}} (8)

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V_{Y}= 45801.13 m/s (11)

This means V_{Y}>V_{X}

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