Ask question

Ask question

All categories

3 years ago

10

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 49.

5 N and it is acting with an effective lever arm of 2.65 cm , what is the torque that the muscle produces on the wrist

1 answer:

UkoKoshka [18]3 years ago

4 0

Answer:

1.1397 Nm

Explanation:

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth.

If the muscle generates a force

$F = 49.5 N$ and $r = 2.65 cm$ , then the torque is equal to $rF$

we see that r = 2.65 cm = 0.0265 m

therefore

torque = 0.0265 x 49.5

= 1.1397 Nm

You might be interested in

Gravity is a force that helps to hold the universe together.true or false

Valentin [98]

Yes, this is a true statement.

gravity is so important.

7 0

3 years ago

Read 2 more answers

What do you need to know to be able to determine how far a projectile travels horizontally?

Effectus [21]

As for me, there are two suitable answers for the question represented above and here is a short explanation why I consider these two to be correct :
D. The horizontal velocity of the projectile and <span>B. The length of time before it lands
</span> $v_x = v*cos(x) [

d_x = v_x(t) + ½at²$ -- this led me to answers! Hope everything is clear! Regards!<span>
</span>

5 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

In the above lightwave the property labeled a determines which characteristic of visible light

kap26 [50]

The light must be either very dim or else non-existent.
We can't see the light wave or the label.

7 0

3 years ago

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang

erma4kov [3.2K]

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$

5 0

3 years ago