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yaroslaw [1]
3 years ago
5

Achargingbullelephantwithamassof5240kgcomesdi- rectly toward you with a speed of 4.55 m/s. You toss a 0.150-kg rubber ball at th

e elephant with a speed of 7.81 m/s. (a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that t
Physics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

Given:

Mass of elephant = 5240 kg

The initial speed of the elephant = 4.55 m/s

Mass of the rubber ball, m, = 0.15 kg

Inital speed of the rubber ball, v = 7.81 m/s

On substitution  in V_{2f} = [\frac{2m_{1} }{m_{1} +m_{2} } ]V_{1f} + [\frac{m_{2}-m_{1}}{m_{1}+m_{2}  } ] v_{2f}

V_{2f} = [\frac{2*5240_{} }{5240_{} +0.15_{} } ](-4.55_{}) + [\frac{0.15_{}-5240_{}}{5240_{}+0.15_{}  } ] (7.81_{})

a) The negatıve sign shows that the ball bounces back in the direction opposite  to the incident

b)  it is clear that the velocity of the ball increases and therefore it is kinetic energy . The ball gains kinetic energy from the elephant.

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False. That's exactly how scientific theories begin

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3 years ago
What force is supplied by a jet’s engines when 7 x 106 Joules of work is required to move the plane down a 450 meter runway?
Contact [7]
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6 0
4 years ago
Which statement correctly describes the amount of force applied by the wall as the student conditions to apply a 250-newton forc
Tatiana [17]

B) The wall pulls with a force of 250 N against the force

Explanation:

We can answer this question by using Newton's third law of motion, which states that:

<em>"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"</em>

In this situation, the student is applying a force of 250 N on the wall. We can identify this force as the action. Therefore, this means (according to Newton's third law) that the wall will also apply a force (the reaction) of equal magnitude (so, 250 N), in the opposite direction to the original force.

Therefore, the correct statement is

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brainly.com/question/11411375

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8 0
3 years ago
You pushes a 25kg box with 40 N of force, but the friction on the crate is 15 N. What is the net
grandymaker [24]

Answer:

25 N  in the direction of your push

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Net force =  40 - 15 = 25 N  

4 0
2 years ago
A Carnot engine operates with a cold reservoir at a temperature of 455 K and a hot reservoir at a temperature of 619 K. What is
Brut [27]

Answer:

Assuming that 4000 J of heat transfer occurs from it the change in entropy will be=2.33j/k

Explanation:

Given data

Th=619 K

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We are going to assume that 4000 J of heat transfer occurs from it, since it was not specified in the question.

we know that the change in entropy is given as

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ΔShot=−Q/Th=−4000J/619K=−6.46J/K

For the cold reservoir,

ΔScold=Q/Tc=4000J/455K=8.79J/K

therefore  the total is

ΔStotal=ΔShot+ΔScold

=(−6.46+8.79)J/K

=2.33j/k

4 0
4 years ago
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