Explanation:
It is given that,
Area of nickel wire, 
Resistance of the wire, R = 2.4 ohms
Initial value of magnetic field, 
Final magnetic field, 
Time, t = 1.12 s
Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :






Induced current in the loop of wire is given by :



So, the induced current in the loop of wire over this time is
. Hence, this is the required solution.
Answer:
792 J
Explanation:
The total energy of the ball is E = U + K where U = potential energy = mgh and K = kinetic energy = 1/2mv²
E = mgh + 1/2mv² where m = mass of ball = 2.0 kg, g = acceleration due to gravity = 9.8 m/s², h = height of building = 20.0 m, v = initial velocity of ball = 20.0 m/s.
So, substituting the values of the variables into E, we have
E = mgh + 1/2mv²
= 2.00 kg × 9.8 m/s² × 20.0 m + 1/2 × 2.00 kg × (20.0 m/s)²
= 392 J + 400 J
= 792 J
Answer:
A. 

B. rev=10.79rev
Explanation:
A.
500rpm

the time is about 2.60s knowing the acceleration is:

so:


B.
The CD stop so the final velocity is vf=0 so:
Total revolutions:


Answer:
FC vector representation

Magnitude of FC

Vector direction FC
degrees: angle that forms FC with the horizontal
Explanation:
Conceptual analysis
Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.
The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.
degrees
To calculate the magnitudes of the forces we apply Coulomb's law:
Equation (1): Magnitude of the electric force of the charge qA over the charge qC
Equation (2)
: Magnitude of the electric force of the charge qB over the charge qC
Known data





Problem development
In the equations (1) and (2) to calculate FAC Y FBC:


Components of the FBC force at x and y:


Components of the resulting force acting on qC:


FC vector representation

Magnitude of FC

Vector direction FC
degrees: angle that forms FC with the horizontal