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11 months ago

A baseball of mass 1.23 kg is thrown at a speed of 65.8 mi/h. What is its kinetic energy?

Physics

1 answer:

frosja888 [35]11 months ago

4 0

Given:

The mass of the ball is

$m=1.23\text{ kg}$

The speed of the ball is

$\begin{gathered} v=65.8\text{ mi/h} \\ \end{gathered}$

Required: calculate the kinetic energy of the baseball

Explanation: to calculate the kinetic energy of a body we will use the formula as

$K.E=\frac{1}{2}mv^2$

first, we convert velocity from mi/h into m/s.

we know that

$1\text{ mi=1609.34 m}$

and

$1\text{ h=3600 sec}$

then the velocity is

$\begin{gathered} v=\frac{65.8\times1602.34\text{ m}}{3600\text{ s}} \\ v=29.29\text{ m/s} \end{gathered}$

now plugging all the values in the above formula, we get

$\begin{gathered} K.E=\frac{1}{2}mv^2 \\ K.E=\frac{1}{2}\times1.23\text{ kg}\times(29.29\text{ m/s})^2 \\ K.E=527.61\text{ J} \end{gathered}$

Thus, the kinetic energy of the baseball is

$527.61\text{ J}$

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1 year ago

If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou

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Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

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We need to calculate the frequency

Case (I),

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$f'=\dfrac{v-v'}{v}f$

Where, v' = speed of observer

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$f'=\dfrac{1540-0.32}{1540}\times4.40$

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Using formula of frequency

$f''=\dfrac{v-v_{s}}{v+v_{s}}\times f$

$f''=\dfrac{1540-0.32}{1540+0.32}\times4.399$

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$\Delta f= f'-f''$

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2 years ago

Two parallel wires are separated by 6.10 cm, each carrying 2.85 A of current in the same direction. (a) What is the magnitude of

Westkost [7]

Answer:

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