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Triss [41]
3 years ago
14

Calculate the moment of inertia for a solid cylinder with a mass of 100g and a radius of 4.0 cm

Physics
2 answers:
Vladimir [108]3 years ago
5 0

Solution here,

mass of cylinder(m)=100 g

radius of cylinder(r)= 4 cm

momoent of inertia(I)=?

we have,

I=mr^2=100×4^2=1600 g/cm^2

yKpoI14uk [10]3 years ago
3 0

Answer:

The moment of inertia for the cylinder is 1600gcm².

Explanation:

In a cylinder with mass m, in g, and radius r, in cm, the moment of inertia is given, in gcm² by the following formula:

I = mr^{2}

In this problem, we have that:

solid cylinder with a mass of 100g and a radius of 4.0 cm. This means that m = 100, r = 4.

So

I = mr^{2} = 100*(4)^{2} = 1600

The moment of inertia for the cylinder is 1600gcm².

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A uniform 40-N board supports two children weighing 500 N and 350 N. If the support is at the center of the board and the 500-N
serg [7]

Answer:

b= 2.14 m

Explanation:

Given that

Weight of the board ,wt = 40 N

Wight of the first children , wt₁=500 N

Weight of the second children ,wt₂ = 350 N

The distance of the 500 N child from center ,a= 1.5 m

lets take distance of the 350 N child from center = b m

Now by taking the moment about the center of the board

We know that moment = Force x Perpendicular distance from the force

wt₁ x a = wt₂ x b

500 x 1.5 = 350 x b

b= 2.14 m

Therefore the distance of the 350 N weight child from the center is 2.14 m.

5 0
3 years ago
Worth 25 points on my exam
I am Lyosha [343]
  • Initial velocity=20m/s
  • Final velocity=0m/s(As the car stops)
  • Acceleration=-8m/s^2
  • Distance=s=26m

We need to verify the thrid equation of kinematics here

\\ \tt\longmapsto v^2-u^2=2as

\\ \tt\longmapsto 20^2=2(-8)s

\\ \tt\longmapsto 400=-16s

\\ \tt\longmapsto s=|400/-16|

\\ \tt\longmapsto s=25m

The squirrel has a good luck ,Car gets stopped just 1m away from the squirrel .

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Irina finds an unlabeled box of fine needles, and wants to determine how thick they are. A standard ruler will not do the job, a
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Answer:

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

Explanation:

The expression which represent the first diffraction minima by a circular aperture is given by d sin\Theta =1.22\lambda--------eqn 1

The angle through which the first minima is diffracted is given by tan\Theta =\frac{y_1}{D}---------eqn 2

As \Theta is very small so we can write sin\Theta =tan\Theta

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It is given that diameter of circular aperture is 14.7 cm so y_1=\frac{14.7}{2}=7.35 \ cm

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d=\frac{1.22\lambda D}{y_1}

d=\frac{1.22\times 640\times 10^{-9}\times 21.7}{7.35\times 10^{-2}}=2305.21\times 10^{-7}m

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olganol [36]

Answer:

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