What current is required in the windings of a long solenoid that has 420 turns uniformly distributed over a length of 0.575 m in
1 answer:
Answer:
The current in the solenoid=0.463 Ampere
Explanation:
Given:
Number of turns , N=420
Length of the solenoid=0.575 m
Magnitude of Magnetic Field,
We know that the magnitude of the magnetic Field inside the solenoid is
Where
is the permeability of free space.
- n is the number of turns per unit length
- i is the current
According to question
Hence the current is calculated.
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Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
