A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns
1 answer:
Answer:
The current is
Explanation:
From the question we are told that
The length of the solenoid is
The radius is
The number of turn is
The magnetic field is
The magnetic field produced is mathematically represented as
making the subject
Where is the permeability of free space with values
substituting values
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Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
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And then we find the final one :
Then equate them together :
So we can get the velocity, which is
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
</span>
And then we find the final one :
Then equate them together :
So we can get the velocity, which is
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
Answer:
The self-induced emf in this inductor is 4.68 mV.
Explanation:
The emf in the inductor is given by:
Where:
dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)
L: is the inductance = 0.260 H
So, the emf is:
Therefore, the self-induced emf in this inductor is 4.68 mV.
I hope it helps you!
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ =
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s