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solmaris [256]
3 years ago
11

A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle

end. You hold the bat at the very tip of the handle end (the knob) and let it swing in simple harmonic motion. What is the bat’s moment of inertia if its period of oscillation is 1.35 seconds?
Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

0.24 kgm²

Explanation:

L = length of the bat = 81.3 cm = 0.813 m

m  = mass of the bat = 0.96 kg

d  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

T  = Period of oscillation = 1.35 sec

I = moment of inertia of the bat

Period of oscillation is given as

T = 2\pi \sqrt{\frac{I}{mgd}}

1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}

I = 0.24 kgm²

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A 2.35-kg rock is released from rest at a height of 21.4 m. Ignore air resistance and determine (a) the kinetic energy at 21.4 m
kvasek [131]

Explanation:

Given that,

The mass of rock, m = 2.35-kg

It was released from rest at a height of 21.4 m.

(a) The kinetic energy is given by : E_k=\dfrac{1}{2}mv^2

As the rock was at rest initially, it means, its kinetic energy is equal to 0.

(b) The gravitational potential energy is given by : E_p=mgh

It can be calculated as :

E_p=2.35\times 9.8\times 21.4\\\\E_p=492.84\ J

(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,

M = 0 J + 492.84 J

M = 492.84 J

Hence, this is the required solution.

6 0
3 years ago
The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it
Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

5 0
3 years ago
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