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Alexxx [7]
3 years ago
11

he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1

03 N with an effective perpendicular lever arm of 2.80 cm, producing an angular acceleration of the forearm of 145 rad/s2. What is the moment of inertia of the boxer's forearm
Physics
1 answer:
meriva3 years ago
5 0

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

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Answer:

temperature at  326.44 K system achieve equilibrium

Explanation:

given data

mass of block of ice = 4.6 kg

temperature = 263 K

thermal contact =  15.7-kg

specific heat of silver  cAg = 233 J/kg-K

initially temperature = 1052 K

to find out

what temperature will the system achieve equilibrium

solution

first we consider final temperature of the system  is T

we know that specific heat of water (C w) = 4186 J(kg K)

and

specific heat of ice ( C i )  = 2030 J/(kg K)

and

latent heat of fusion of ice ( Lf ) = 3.33 × 10^{5} J/kg

and we know that system is insulated

so  heat lost by silver = heat generated by ice    .................1

so we can say

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )  

put here value we get

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )

15.7  × 233 × ( 1052 - T ) = 4.6 × 2030 × 10 + 4.6 × 3.33 × 10^{5} + 4.6 × 4286 × ( T - 273 )

solve we get

T =  326.44 K

so temperature at  326.44 K system achieve equilibrium

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Answer:

2.07

Explanation:

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Squaring both sides:

n^2 * L2^2 = L1^2 + L2^2……….12

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8 0
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