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klio [65]
1 year ago
10

What is the motional kinetic energy of a 25 kg object moving at a speed of 10 m/s?

Physics
1 answer:
Soloha48 [4]1 year ago
5 0

1250kgm²/s is the motional kinetic energy of a 25kg object moving at a speed of 10m/s

Kinetic energy of an object is defined as the energy which is possessed when that is  in motion. It is the energy of the kinetic mass of an object. Kinetic energy is never negative and is a scalar quantity. That is, it shows only size, not orientation.

Given to us

Mass of the object, m=25kg

Velocity of the object, v=10m/s  

K.E=1/2x25x10²

 =1250

Kinetic energy is directly proportional to the mass and velocity squared (K.E.) of an object. =1/2xMxV². If the mass is in kilograms and the velocity is in meters/second, then the kinetic energy is in kilograms - meters squared/second.

Learn ore about Kinetic energy here brainly.com/question/25959744

#SPJ9

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Equations. Mass and the square of the velocity are directly related to translational kinetic energy. The difference between the end and starting kinetic energies is known as change in kinetic energy.

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8 0
1 year ago
I need help with egg drop assignments any ideas?
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Answer:

make a parachute out of the bag connecting to a bowl made out of paper filled with cotton balls. then put the egg in the bowl.

Explanation:

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3 years ago
What Is The frequency of an X Ray That had A Wavelength of 1.5* 10^-9
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When making a turn, do not have the steering wheel turned in the direction of the turn before beginning the turning maneuver.a)
RideAnS [48]

Answer:

a) True.

Explanation:

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7 0
4 years ago
At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw
allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

f_i = f_0 \frac{c}{c+v}

Where

f_h= Approaching velocities

f_i= Receding velocities

c = Speed of sound

v = Emitter speed

And

f_h = f_0 \frac{c}{c+v}

Therefore using the values given we can find the velocity through,

\frac{f_h}{f_0}=\frac{c-v}{c+v}

v = c(\frac{f_h-f_i}{f_h+f_i})

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

v = 353(\frac{2.4-1}{2.4+1})

v = 145.35m/s

Therefore the cars goes to 145.3m/s

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