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Alexandra [31]
3 years ago
12

The triceps muscle in the back of the upper arm extends the forearm.

Physics
1 answer:
iren [92.7K]3 years ago
8 0

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

\tau = F \times r

Where,

F = Force

r = Distance

Replacing,

\tau = 2*10^3*0.03

\tau = 60N\cdot m

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

\tau = I \alpha

I = Moment of inertia

\alpha = Angular acceleration

Replacing with our values we have that

I = \frac{\tau}{\alpha}

I = \frac{60}{120}

I = 0.5kg\cdot m^2

Therefore the value of moment of inertia is 0.5kg\cdot m^2

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A racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction). It hits the wall of th
zheka24 [161]

The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m  = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Learn more on momentum here: brainly.com/question/402617

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If the moon were 2 times closer to earth than it is now, the gravitational force between earth and the moon would be
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3 years ago
Diffusion and osmosis are forms of passive transport.<br><br> True<br> False
pishuonlain [190]

Answer:

True. Diffusion and osmosis are forms of passive transport.

Explanation:

In diffusion, particles move from an area of higher concentration to one of lower concentration until equilibrium is reached.

In osmosis, a semipermeable membrane is present, so only the solvent molecules are free to move to equalize concentration.

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During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25
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Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

dQ=0

dU=-dW

We need to calculate the number of moles and specific heat

Using formula of heat

dU=nC_{v}dT

nC_{v}=\dfrac{dU}{dT}

Put the value into the formula

nC_{v}=\dfrac{-100}{5}

nC_{v}=-20\ J/K

We need to calculate the heat

Using formula of heat

dQ=nC_{v}(dT_{1})+dW_{1}

Put the value into the formula

dQ=-20\times5+25

dQ=-75\ J

We need to calculate the heat capacity for the second process

Using formula of heat

dQ=nC_{v}(dT_{1})

Put the value into the formula

-75=nC_{v}\times(-5)

nC_{v}=\dfrac{-75}{-5}

nC_{v}=15\ J/K

Hence, The heat capacity for the second process is 15 J/K.

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