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RUDIKE [14]
3 years ago
15

According to the conservation of momentum, if a small football player collides with a much larger football player, which of the

following statements will be true? a The larger football player will move, but because of his larger mass, his velocity will be much slower than the smaller player b The smaller football player will be stopped completely, his mass being not enough to move the larger one c The larger football player will have to start moving towards the smaller one as it collides if there is to be any effect d The smaller football player will experience a larger force than the larger football player
Physics
1 answer:
Virty [35]3 years ago
8 0

Answer:

a The larger football player will move, but because of his larger mass, his velocity will be much slower than the smaller player

Explanation:

When the collision takes two equal and opposite forces are created at the point of collision . The force created on the bigger mass will force it to accelerate and the force on moving smaller mass will force it to slow down .

Because of bigger mass , acceleration on bigger mass will be less and hence its velocity will be less .

Hence option a is the right answer.

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20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu
Anna71 [15]

Answer:

B. V_{f}= 10\,cubic\,inches

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

PV=nRT

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

P_{i}V_{i}=n_{i}RT_{i} (1)

P_{f}V_{f}=n_{f}RT_{f} (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because T_{i}=T_{f}, n_{i}=n_{f} and R is an universal constant:

P_{i}V_{i}= P_{f}V_{f}, solving for V_{f}

V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}

V_{f}= 10 cubic\,inches

6 0
3 years ago
Please Help............................
vlada-n [284]
Protons
electrons are negatively charged
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7 0
3 years ago
Read 2 more answers
Is friction and pushing similar ????
olya-2409 [2.1K]

Answer:

Yes, they are.

Explanation:

3 0
3 years ago
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Determine the gain in the potential energy when a 8.0 kg box is raised 17.2 m.
Marysya12 [62]

Answer:

<h2>The answer is 1376 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 8 × 10 × 17.2

We have the final answer as

<h3>1376 J</h3>

Hope this helps you

6 0
3 years ago
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What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}&#10; 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
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