In a transverse wave, ____________________ is measured from crest to crest or from trough to trough.

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b

taurus [48]

Answer:

I = 0.483 kgm^2

Explanation:

To know what is the moment of inertia I of the boxer's forearm you use the following formula:

$\tau=I\alpha$ (1)

τ: torque exerted by the forearm

I: moment of inertia

α: angular acceleration = 125 rad/s^2

You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

$\tau=Fr=(1.95*10^3N)(0.031m)=60.45J$

Next, you replace this value of τ in the equation (1) and solve for I:

$I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2$

hence, the moment of inertia of the forearm is 0.483 kgm^2

8 0

3 years ago

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is:

$a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)$

$a_{t} = -1.075\, \frac{m}{s^{2}}$

The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) If we know that $\omega_{o} = 104.720\,\frac{rad}{s}$ and $\omega = 26.180\,\frac{rad}{s}$ , then the time needed is:

$26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}$

$5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}$

$t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5}$

$t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5} }$

$t \approx 3.963\,s$

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

8 0

3 years ago

Where is parallel circuits used?

suter [353]

Answer:

if one bulb burns out, the other bulbs in the fixture continue to operate. Other uses include an electronic OR gate, where two switches are in a parallel circuit: one of the switches must be closed for the circuit to function.

8 0

3 years ago

What type of wave is shown below?

Elena-2011 [213]

Answer:

B. Transverse wave.

Explanation:

Because it has troughs and crests.

5 0

3 years ago

How does the heat from sun reach earth

kakasveta [241]

Answer:

The sun heats the earth through radiation. Since there is no medium (like the gas in our atmosphere) in space, radiation is the primary way that heat travels in space. When the heat reaches the earth it warms the molecules of the atmosphere, and they warm other molecules and so on.

3 0

3 years ago