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>
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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.
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Answer:
0.46786 W
Explanation:
The solution is in the attached file below
Answer:
average speed and velocity are 0.5m/s (squared)
Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
given that initial speed of the car is
now after travelling the distance d = 1.8 * 10^1 m the car will stop
so here we can use kinematics to find the acceleration of car
here we have
net force applied due to brakes of car is given by Newton's II law
here we have
mass = 1.2 * 10^3 kg
now we can say
So the force applied due to brakes is given as above
