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emmasim [6.3K]
1 year ago
12

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]1 year 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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Answer:

It's displacement would be 25 km.

Explanation:

That is because you would subtract 10 from 35 to get the distance away from the starting point.

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Answer:

2144.66 cm³

Explanation:

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3 years ago
an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20
Schach [20]
Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
1/5 = m

at 20 km/hr

KE = (1/2)(1/5)(20)^2
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What units are used to measure mass and weight?
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RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s
Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, R=50\ K\Omega=5\times 10^4\ \Omega

Capacitance, C=2.1\ \mu F=2.1\times 10^{-6}\ F

We need to find the expected time constant for this RC circuit. It can be calculated as :

\tau=R\times C

\tau=5\times 10^4\times 2.1\times 10^{-6}

\tau=0.105\ s

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

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