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Aleksandr [31]
3 years ago
6

A skater has a moment of inertia of 105.0 kg.m^2 when his arms are outstretched and a moment of inertia of 70.0 kg.m^2 when his

arms are tucked in close to his chest. If he starts to spin at an angular speed of 80.0 rpm (revolutions per minute) with his arms outstretched, what will his angular speed be when they are tucked in?
Physics
1 answer:
kati45 [8]3 years ago
4 0

Answer:

120 rpm

Explanation:

I1 = 105 kgm^2, I2 = 70 kgm^2

f1 = 80 rpm, f2 = ?

Let the angular speed be f2 when his arms are tucked.

If no external torque is applied, then the angular momentum remains constant.

L1 = L2

I1 w1 = I2 w2

I1 x 2 x 3.14 x f1 = I2 x 2 x 3.14 x f2

105 x 80 = 70 x f2

f2 = 120 rpm

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Answer:

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3 years ago
the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?
Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

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a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

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P = 1000*9.81*0.04

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P = F/A

F = P*A

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The Answer Is A 
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