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3 years ago

a 10kg ball is thrown into the air. it is going 3m/s when thrown. How much potential energy will it have at the top?

Physics

1 answer:

Alexandra [31]3 years ago

7 0

Answer:

45 J

Explanation:

Assuming the level at which the ball is thrown upwards is the ground level,

We can use the equations of motion to obtain the maximum height covered by the ball and then calculate the potential energy

u = initial velocity of the ball = 3 m/s

h = y = vertical distance covered by the ball = ?

v = final velocity of the ball at the maximum height = 0 m/s

g = acceleration due to gravity = -9.8 m/s²

v² = u² + 2ay

0 = 3² + 2(-9.8)(y)

19.6y = 9

y = (9/19.6)

y = 0.459 m

The potential energy the ball will have at the top of its motion = mgh

mgh = (10)(9.8)(0.459) = 45 J

Hope this Helps!!!

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To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

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$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

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3 years ago

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Answer:

Explanation:

Motor rating is given in horsepower (hp), it will be converted in watt (W).

Standard to install circuit breaker for an electric circuit is usually 20% ~ 25% more than the Rated Current of the circuit

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Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

So, to find the maximum capacity of the circuit breaker, rated current of the motor will be multiplied by 1.2 ~ 1.25

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<em>{if 208V is phase to neutral voltage, then use following formula:</em>

<em> Rated Power (watt) = 3 . V.I. cosФ}</em>

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