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Pavlova-9 [17]
3 years ago
11

A baseball is hit into the air with a velocity of 27 m/s. How high does it go?

Physics
1 answer:
Marizza181 [45]3 years ago
8 0

Answer:

Explanation:

If a baseball is hit into the air with a velocity of 27 m/s, we want to determine the maximum height of the ball. Using the projecile formula;

Max height H = u²/2g

u is the initial velocity of the body = 27m/s

g is the acceleration due to gravity = 9.81m/s²

H = 27²/2(9.81)

H = 729/19.62

H = 37.16m

Hence the ball went 37.16m high

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Transformer contains two coils: primary and secondary. They allow change of voltage to lower or higher value. In first case we have step-down and in second case we have step-up transformer.
Formula used for transformer is:\frac{N_{1} }{N_{2}} = \frac{V_{1}}{V_{2}}

Where:N1 = number of turns on primary coilN2 = number of turns on secondary coilV1 = voltage on primary coilV2 = voltage on secondary coil
In a step-down transformer primary coil has more turns than secondary coil. So the ratio 1:38 means that for each turn on secondary coil we have 38 turns on primary coil.
We can solve the equation for V2:V_{2} =  \frac{ V_{1}* N_{2}  }{ N_{1} }  \\  V_{2} =  \frac{ 120* x  }{ 38x} } \\ V_{2} = 3.16V

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b) 0.0564 s

Explanation:

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m = 2100 kg

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vf = 0 ......after collision to stop

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the answer is

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