The strength of the magnetic field is 
Explanation:
According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:
(1)
where
N = 505 is the number of turns in the coil
is the change in magnetic flux through the coil
is the time interval
The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:
where
B is the strength of the magnetic field
A is the area of the coil
is the angle between the normal to the coil and the field
The area of the coil can be written as
where
is its radius
Substituting everything into eq.(1) and solving for B, we find:
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Answer:
Magnetic field, B = 0.004 mT
Explanation:
It is given that,
Charge, 
Mass of charge particle, 
Speed, 
Acceleration, 
We need to find the minimum magnetic field that would produce such an acceleration. So,
For minimum magnetic field,
B = 0.004 T
or
B = 4 mT
So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.
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.
Answer:
Explanation:
For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.
X axis
-Aₓ - f_e +T = 0 (1)
Y axis
N₁ - W_y = 0 ( 2)
let's use trigonometry for the weight components
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
We write the diagram for the second body.
Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards
W₂ -T = 0 (3)
we add the equations is 1 and 3
- W₁ sin θ - μ N₁ + W₂ = 0
from equation 2
N₁ = W₁ cos θ
we substitute
-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0
W₂ = m₁ g (without ea - very expensive)
This is the smallest value that supports the equilibrium system
