) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tigh
1 answer:
You might be interested in
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
Answer:
The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.
Explanation:
Given that,
Mass of bucket = 54 kg
Radius = 0.050 m
We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder
Using formula of torque
Where, m = mass
g = acceleration due to gravity
r = radius
Put the value into the formula
Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.