Answer:
33,458.71 turns
Explanation:
Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter
μ₀ = Permeability of free space = 4 π × 10 ⁻⁷
Solution:
We have B = μ₀ × n × I
⇒ n = B/ (μ₀ × I)
n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)
n = 90,428.94 turn/m
No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37
= 33,458.71 turns
Answer:
Fr = 48 [N] forward.
Explanation:
Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.
∑F = Fr
![F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N]](https://tex.z-dn.net/?f=F_%7Bengine%7D-F_%7Bbrake%7D%20%3DF_%7Br%7D%5C%5CF_%7Br%7D%3D79-31%5C%5CF_%7Br%7D%3D48%5BN%5D)
The movement remains forward, since the force produced by the movement is greater than the braking force.
Answer:
The first minimum would be observed at 41.57°
Explanation:
v = 340m/s = speed of sound
f = 610Hz
d = 0.840m
λ = ?
Mλ = wsinθ
m = mth order minima
λ = wavelength incident on the single slit
θ = angular position of the mth minima
But, λ = v / f
λ = 340 / 610 = 0.557m
θ = sin⁻(mλ/d)
θ = sin⁻ [(1 * 0.557) / 0.840]
θ = sin⁻ 0.6635
θ = 41.57°
The first minimum would be observed at 41.57°
Answer:
τ=0.060 N.m
Explanation:
By kinematics:
Solving for α:
where ωo = 600*2*π/60; ωf = 0; t=10s
The sum of torque is:
