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Sedaia [141]
3 years ago
9

A 72.0-kg object hits the ground at a velocity of 79.0 m/s. Neglecting air resistance, which relationship allows you to calculat

e the object’s potential energy before the fall? A. PE = mgh B. (PE)beginning = (KE)end C. Mechanical Energy = PE + KE D. KE = 1/2 mv2
Physics
1 answer:
kondaur [170]3 years ago
8 0
<h3>Answer:</h3>

B. (PE)beginning = (KE)end

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of an object is 72.0 kg
  • Velocity of the body before it hits the ground is 79.0 m/s

We are required to determine the relationship between the potential energy and kinetic energy before the fall.

  • When an object is at the highest point, it has maximum potential energy and minimum kinetic energy.
  • This is because potential energy is directly proportional to the height of an object above the earth's surface.
  • On the other hand, when an object attains the highest speed it has maximum kinetic energy and minimum kinetic energy.

In this case;

  • The velocity of the object when hitting the ground is maximum and thus the object will have maxim,um kinetic energy.
  • As the object falls towards the ground the potential energy is being converted to kinetic energy.
  • Therefore, the potential energy at the beginning will be equal to the kinetic energy at the end when the object is on the ground.
  • We can therefore, conclude that, (PE)beginning = (KE)end
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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock
pashok25 [27]

Answer:

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, a=2g=2\times 9.81=19.62\ m.s^{-2}

time for which the rocket accelerates, t_a=100\ s

<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

u= initial velocity of the rocket at the launch =0

s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

where:

g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

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Nastasia [14]

Answer:

<u><em>on  flow properties and free-flowing and cohesive. </em></u>

Explanation:

the power Free flowing powders do not cling together, as cohesive powders stick to each other and form that do not disperse well during mixing

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docker41 [41]
I would say mass, and weight.
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posledela
The ball does not have D. Radiant energy.
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I would say that the answer is A.
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