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Lelu [443]
3 years ago
13

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

ds, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 29 m/s. The masses of the two objects are 3.2 and 7.1 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially
Physics
1 answer:
Bess [88]3 years ago
7 0

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p =  momentum.

m =  mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

v2 = velocity of the smaller object.

v = final velocity of the two objects after the collision.

Solving the equation for the final velocity:

(m1 · v1 + m2 · v2)/ (m1 + m2) = v

a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

<u>v = 20 m/s</u>

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

<u>v = 9.0 m/s</u>

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Answer:

The answer is

<h2>270 m</h2>

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time = 15 s

So the distance is

distance = 18 × 15

We have the final answer as

<h3>270 m</h3>

Hope this helps you

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<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

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Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

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Substituting

                   F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m

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Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

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Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

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