Answer:

Explanation:
As we know that magnetic flux through the coil is given by

now by Faraday law we know that rate of change in magnetic flux is equal to the EMF induced in the coil
so we have


now we have


now we have



now at t = 0.10 s

In would say that you may have water in your brakes which may have gotten in the brake lines or in the brake discs so that could cause the brakes to malfunction due to driving through the pools of water so the brakes should be examined as soon as possible.
The work done on the barbell is -165.62 Nm.
Explanation:
Work done on any object is the measure of force required to move that object from one position to another. So it is determined by the product of force acting on the object with the displacement of the object.
In the present problem, the displacement of the object on acting of force is given as 1.3 m. And the weight of the object which is a barbel is given as 13 kg. As the work is to lift the object from the ground, so the acceleration due to gravity will be acting on the object. In other words, the force applied on the object to lift it should be in opposite direction to the acting of acceleration due to gravity.
Thus, 
Now, the force is -127.4 N and the displacement is 1.3 m.
So, 

So, the work done on the barbell is -165.62 Nm.
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>