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Bas_tet [7]
3 years ago
12

What is the difference between potential and kinetic types of energy? 1 pois​

Physics
2 answers:
GREYUIT [131]3 years ago
5 0
Answer: The main difference between potential and kinetic energy is that one is the energy of what can be and one is the energy of what is. In other words, potential energy is stationary, with stored energy to be released; kinetic energy is energy in motion, actively using energy for movement.
Explanation: hoped the helped :)
Tasya [4]3 years ago
3 0

Answer:

Potential energy is energy that is stored in an object or system. It remains unaffected by the environment outside of the object or system. Kinetic energy is the energy of an object or a system's particles in motion.

Explanation:

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How are temperature, pressure, and volume related dealing with the behavior of a gas
fredd [130]
Pressure and volume of a gas are inversely related. As one goes up, the other goes down, and vice-versa. 
7 0
4 years ago
Calculate the object's velocity as shown on the position-time graph,
Cloud [144]

Answer:

10 m/s

Explanation:

The following data were obtained from the question:

Initial Displacement (d1) = 10 m

Final Displacement (d2) = 60 m

Initial time (t1) = 0 s

Final time (t2) = 5 s

Velocity (v) =.?

Next, we shall determine the change in displacement of the object and likewise the change in time.

This can be obtained as follow:

Initial Displacement (d1) = 10 m

Final Displacement (d2) = 60 m

Change in displacement (Δd) = d2 – d1

Change in displacement (Δd) = 60 – 10

Change in displacement (Δd) = 50 m

Initial time (t1) = 0 s

Final time (t2) = 5 s

Change in time (Δt) = t2 – t1

Change in time (Δt) = 5 – 0

Change in time (Δt) = 5 s

Finally, we shall shall calculate the velocity of the object as illustrated below:

Change in displacement (Δd) = 50 m

Change in time (Δt) = 5 s

Velocity (v) =.?

v = Δd/Δt

v = 50/5

v = 10 m/s

Therefore, the velocity of the object is 10 m/s.

4 0
3 years ago
Question 1 (1 point)
MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work W done by the frictional force is

W= Fd.

Now, F = 60N and d =10m; therefore,

W= (60N)(10m)

\boxed{W = 600J.}

Hence, the work done by friction is 660J.

7 0
3 years ago
Which of the following is a good step in learning how to manage time?
Luden [163]
I think is c because your seeing how your using you’re time.
8 0
3 years ago
Read 2 more answers
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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