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sp2606 [1]
4 years ago
5

A 1530 kg car moving south at 12.1 m/s collides with a 2560 kg car moving north. The cars stick together and move as a unit afte

r the collision at a velocity of 5.48 m/s to the north. Find the velocity of the 2560 kg car before the collision. Assume that North is positive. Answer in units of m/s.
Physics
1 answer:
sveticcg [70]4 years ago
8 0

Answer:

16 m/s north

Explanation:

According to the law of momentum conservation, total momentum before and after must be the same

Total momentum after the collision as both cars stick together as 1 unit traveling at speed of 5.48m/s to the north

(m + M)V = (1530 + 2560)*5.48 = 22413 kgm/s

So the total momentum before must also be 22413, which is made of

mv_1 + Mv_2 = 22413

1530*(-12.1) + 2560v_2 = 22413

2560v_2 = 22413 + 18513 = 40926

v_2 = 40926 / 2560 \approx 16 m/s north

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What are the messing forces that would make the object be in equilibrium?
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Answer:

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Explanation:

For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.

1. Determination of A and B.

Forward forces = Backward forces

A + 10 + B = 25 + 25

A + 10 + B = 50

Collect like terms

A + B = 50 – 10

A + B = 40

Assume A and B to be equal. Thus, A is 20 N and B is 20 N.

2. Determination of C

Upward forces = Downward forces

C + 112 = 20 + 100

C + 112 = 120

Collect like terms

C = 120 – 112

C = 8 N

Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.

6 0
3 years ago
A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?
Dmitriy789 [7]

Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

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1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

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3 years ago
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3 years ago
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Can't say for sure but I think it's a) or c)
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3 years ago
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