If you double the number of turns and double the length, the value N/L remains the same and hence the magnitude of the magnetic field stays the same.
A solenoid is a device comprised of a coil of wire, the housing and a moveable plunger (armature). When an electrical current is introduced, a magnetic field forms around the coil which draws the plunger in.
The magnitude of the magnetic field in an infinite (L>>R) solenoid, is proportional to:
I*N/L
Where
I = current
N = number of turns
L = length of wire
So, if you double the number of turns and double the length, the value N/L remains the same and hence the magnitude of the magnetic field stays the same.
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The answer is kinetic energy
Answer:
Anita 's power rating during this portion of the climb is 1568 Watts.
Explanation:
Given that,
Mass of the body, m = 800 kg
Height, h = 20 m
Time, t = 100 s
We need to find the Anita 's power rating during this portion of the climb. Power rating of an object is given by the work done per unit time. It is given by :
P = 1568 Watts
So, Anita 's power rating during this portion of the climb is 1568 Watts. Hence, this is the required solution.
Answer:
56 kg
Explanation:
The change in potential energy of the man is given by:
where
m is the man's mass
g is the gravitational acceleration
is the change in height of the man
In this problem, we have:
is the gain in potential energy
g = 9.8 m/s^2 is the gravitational acceleration
is the change in height
Re-arranging the equation and substituting the numbers, we find the mass:
The shaft is 78m approximately deep. The correct option is D which is 78 meters
<h3>
What are Sound Waves ?</h3>
Sound waves are longitudinal. That is, the direction of the waves is parallel to the direction of its propagation of particles.
Given that a boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seconds before he hears the "plunk" of the stone hitting the bottom of the shaft.
- The speed of the sound V = 330m/s
- The time it takes the sound to reach the top = t
Speed V = distance / time
330 = h/t
Make t the subject of formula
t = h/330
As the stone is dropped, initial velocity = 0, Using the formula
h = 1/2gT²
But T = 4 - t
T = 4 approximately
Substitute all the parameters
h = 1/2 × 9.8 × (4)²
h = 4.9 × (16)
h = 78.4m
Therefore, the shaft is 78m approximately deep. The correct option is D which is 78 meters
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