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avanturin [10]
3 years ago
10

What is the formula for kinetic energy

Physics
2 answers:
coldgirl [10]3 years ago
8 0

Answer:

The formula for kinetic energy is K.E. = 1/2 mv 2 , where "m" stands for mass and "v" stands for velocity. Kinetic energy is typically measured in units of Joules, and 1 Joule is equal to 1 kilogram-meters squared per second squared.

Explanation:

gregori [183]3 years ago
7 0

Answer:

KE = 0.5 x mv^2

Explanation:

KE = Kinetic energy

m = mass

v = velocity

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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
Q 36 - URGENT HELP PLS<br> WILL BE MARKED AS BRAINLIEST.
Sliva [168]
The answer is B.

This is because you add up all of the times (1.44s+1.70s+1.58s+1.76s) and you get 6.48 then you divide 6.48 by 4 to get the average of the times. Now you get the distance (200m) and because speed=distance/time you divide 200m/1.62s to get 123m/s. I hope this made sense :)
5 0
3 years ago
If the pressure of gas is doubled and its absolute temperature is quadrupled, the volume is what factor times the original?
Lapatulllka [165]

Answer:

Volume will increase by factor 2

So option (A) will be correct answer  

Explanation:

Let initially the volume is V pressure is P and temperature is T

According to ideal gas equation PV=nRT, here n is number of moles and R is gas constant

So V=\frac{nRT}{P}....................eqn 1

Now pressure is doubled and temperature is quadrupled

So new volume V_{new}=\frac{nR4T}{2P}=\frac{2nRT}{P} ........eqn 2

Now comparing eqn 1 nad eqn 2

V_{new}=2V

So volume will increase by factor 2

So option (A) will be correct answer

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Please do number 11 need this question bad
ArbitrLikvidat [17]

Gas always expands or contracts to exactly fill whatever you put it in.

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