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
and
, then the torque is equal to 
we see that r = 2.65 cm = 0.0265 m
therefore
torque = 0.0265 x 49.5
= 1.1397 Nm
Answer:
38.87 m/s
Explanation:
Given that the ball is dropped from a height = 77 m
u = 0 m/s
s = 77 m
a = g = 9.81 m/s²
Applying the expression as:

Applying values as:

<u>The speed with which the ball hit the ground = 38.87 m/s</u>
Answer:

Explanation:
Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.
- m= 150 kg
- g= 9.8 m/s²
- h= 20 m
Substitute the values into the formula.

Multiply the three numbers and their units together.


Convert the units.
1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

The crate has <u>29,400 Joules</u> of potential energy.
Answer:
The value is 
Explanation:
From the question we are told that
The radius of the inner conductor is 
The radius of the outer conductor is 
The potential at the outer conductor is 
Generally the capacitance per length of the capacitor like set up of the two conductors is
![C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}](https://tex.z-dn.net/?f=C%3D%20%5Cfrac%7B2%20%2A%20%5Cpi%20%2A%20%5Cepsilon_o%20%7D%7B%20ln%20%5B%5Cfrac%7Br_2%7D%7Br_1%7D%20%5D%7D)
Here
is the permitivity of free space with value 
=> ![C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}](https://tex.z-dn.net/?f=C%3D%20%5Cfrac%7B2%20%2A%20%203.142%20%20%2A%208.85%2A10%5E%7B-12%7D%20%20%7D%7B%20ln%20%5B%5Cfrac%7B0.003%7D%7B0.001%7D%20%5D%7D)
=> 
Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge
Generally the line charge density of the outer conductor is mathematically represented as

=> 
=> 
Generally the surface charge density is mathematically represented as
here 
=> 
=> 