Answer:
Explanation:
Induced EMF in the coil is given by the equation
so we have
also we know that rate of change in current in solenoid is given as
so induced EMF of coil is given as
now induced current in the coil will be given as
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
Answer:
V
I and II
III and IV
Explanation:
The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.
Taking always east as positive direction, and labelling
u the initial velocity
v the final velocity
m = 1000 kg the mass (which is always equal)
We find:
(i)
u = 25 m/s
v = 0
(II)
u = 25 m/s
v = 0
(III)
In this case,
F = 2000 N is the force
is the time
So the magnitude of the impulse is
(IV)
F = 2000 N is the force
is the time
So the magnitude of the impulse is
(V)
u = 25 m/s
v = -25 m/s
So the ranking from largest to smallest is:
V
I and II
III and IV
Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.
