4hz is the answer i think atleast
The self-inductance of the solenoid is 8.25 mH.
The given parameters;
- <em>number of turns, N = 1700 turn</em>
- <em>length of the solenoid, l = 55 cm = 0.55 m</em>
- <em>diameter of solenoid, d = 4 cm</em>
- <em>radius of the solenoid, r = 2 cm = 0.02 m</em>
<em />
The area of the solenoid is calculated as follows;
The self-inductance of the solenoid is calculated as follows;
Thus, the self-inductance of the solenoid is 8.25 mH.
Learn more here:brainly.com/question/15612168
The velocity of the photo electron is 6.11 × 105 ms
Given :
Supplied energy, Es = 5.3 x 10-19 J
Minimum energy of the electron to escape from the metal, E. = 3.6 x 10-19 J
To Find :
Velocity of photoelectron
Solution : The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So
5.3 × 10^-19 J = 3.6 × 10^-19 J + K
K = 5.3 x 10^-19 - 3.6 x 10^-19
K = 1.7 × 10^-19 J
The formula of kinetic energy is given by:
K = 1/2 mv^2
v = √2K/m
= <u>√2 x 1.7 x 10^-19</u>
√ 9.1 x 10^-31
v = 6.11 x 10^5 m/s
So, the velocity of the photo electron is 6.11 x 10^5 m/s
Learn more about Velocity here:
brainly.com/question/15585270
#SPJ4
<h2>
Answer: 1000 J</h2>
The Work done by a Force refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:
Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
(1)
When they are not parallel, both directions form an angle, let's call it . In that case the expression to calculate the Work is:
(2)
For example, in order to push the 200 N box across the floor, you have to apply a force along the distance to overcome the resistance of the weight of the box (its 200 N).
In this case both <u>(the force and the distance in the path) are parallel</u>, so the work performed is the product of the force exerted to push the box by the distance traveled . as shown in equation (1).
Hence:
>>>>This is the work