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solmaris [256]
2 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]2 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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What is the factor involved in increasing an object’s inertia?
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Inertia= inactivity
The factor involved in increasing inertia would be "mass". The more mass something has, the more inertia. 
7 0
2 years ago
Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. The electric field at a point P, which
krok68 [10]

Answer:

E_{max}=41666.66\ N/C

Explanation:

Given that,

The radius of sphere, r = 0.3 m

Distance from the center of the sphere to the point P, x = 0.5 m

Electric field at point P, E_P=15000\ N/C (radially outward)

The maximum electric field is at the surface of the sphere. We know that the electric field is inversely proportional to the distance. So,

\dfrac{E_{max}}{E_p}=\dfrac{0.5^2}{0.3^2}

\dfrac{E_{max}}{15000}=\dfrac{0.5^2}{0.3^2}

{E_{max}}=\dfrac{0.5^2}{0.3^2}\times 15000

E_{max}=41666.66\ N/C

So, the magnitude of the electric field due to this sphere is 41666.66 N/C. Hence, this is the required solution.

6 0
1 year ago
A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
12*8A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 13 feet. The ball is started in
max2010maxim [7]

Answer:

See attached pictures.

Explanation:

See attachments for explanation.

6 0
1 year ago
A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an
inna [77]

Answer:

It will take 4 sec rock to comes its original point

Explanation:

It is given that the rock comes to its original point

So displacement S = 0 m

Initial velocity u = 19.6 m/sec

Acceleration due to gravity g=9.8m/sec^2

According to second equation of motion h=ut+\frac{1}{2}gt^2

0=19.6\times t+\frac{1}{2}\times 9.8t^2

19.6=4.9t

t = 4 sec

3 0
2 years ago
Read 2 more answers
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