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Ugo [173]
4 years ago
15

An 85-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical

elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Physics
1 answer:
Bas_tet [7]4 years ago
5 0

Answer:

The change in gravitational potential energy of the climber-Earth system,\Delta E=4.49\times 10^5\ J

Explanation:

Given that,

Mass of the hiker, m = 85 kg

Time, t = 2 h

Vertical elevation of the climber, h = 540 m

We need to find the change in gravitational potential energy of the climber-Earth system. We know that due to change in position of an object, gravitational potential energy occurs. It is given by :

\Delta E=mgh\\\\\Delta E=85\times 9.8\times 540\\\\\Delta E=4.49\times 10^5\ J

So, the change in gravitational potential energy of the climber-Earth system is 4.49\times 10^5\ J. Hence, this is the required solution.  

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The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

t=68.94 m

8 0
3 years ago
HELPPPPPPP PLEASE!!! does anybody have the answer key to edge. physics lab report : mechanical equivalent to heat? one page of d
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Answer:

the mechanical equivalent of heat states that motion and heat are mutually interchangeable and that in every case

Explanation:

6 0
3 years ago
Read 2 more answers
Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on
lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0
4 years ago
A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. How much energy is lost? Group of answer choices 0.5
bogdanovich [222]

Answer: A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. Energy loss will be 0.29J

Explanation: To find the correct answer, we have to know more about the Gravitational potential energy.

<h3>What is gravitational potential energy?</h3>
  • The energy possessed by a body by virtue of its position in gravitational field of earth is called gravitational potential energy.
  • The gravitational potential energy of a body at a height h with respect to the height h will be,

                                          U=mgh

  • Expression for gravitational potential energy loss will be,

                                        E=U_i-U_f

<h3>How to solve the problem?</h3>
  • The total energy before the ball dropped will be,

                 U_i=mgh_i=60*10^-3kg*9.8m/s^2*2m=1.176 J

  • The total energy after when the ball rebounds to 1.5m will be,

                 U_f=mgh_f=60*10^-3kg*9.8m/s^2*1.5m=0.882J

  • The total energy loss will be,

                E=1.176-0.882=0.294J

Thus, we can conclude that, the energy loss will be,0.294J.

Learn more about the gravitational potential energy here:

brainly.com/question/28044692

#SPJ4

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2 years ago
A neutral object has:
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Answer:

c. Only neutrons.

hope it helps :)

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2 years ago
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