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MrRissso [65]
3 years ago
5

A 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?(unit=W)PLEASE H

ELP
Physics
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .

Explanation:

Dont report my answer please

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What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m
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Answer:

Velocity of a proton, v=1.7\times 10^6\ m/s    

Explanation:

It is given that,

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Using the conservation of energy as :

\dfrac{1}{2}mv^2=qV

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v=\sqrt{\dfrac{2qV}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}

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