Answer:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
It evaporates into a vapor
~~~hope this helps~~~
~~have a beautiful day~~
~davatar~
To explain, I will use the equations for kinetic and potential energy:

<h3>Potential energy </h3>
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>
Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>
Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
Now, we can find the drift speed:
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
14-needle heading west
15-the strength of the current and the distance