Answer:
0.00479 volts
Explanation:
From Faraday's law of electromagnetic induction, the induced emf equals the change in magnitude flux (magnetic field strength multiplied by the area = BA) divided by the time change
Therefore we have the equation
EMF = BA÷t
Since area A = 2πr
EMF = Bπr²÷t
B = 5.5×10^(-5)
Velocity = 100km/h = 27.7778m/s
r = 1m, t = r÷V = 0.036
EMF = Bπr²÷t = (5.5×10^(-5) x π x (1)²)÷0.036 = 0.00479 Volts
Explanation:
From the given figure,
Mass of ball A is 4.9 kg and its initial speed is 8 m/s.
Mass of ball B is 1.9 kg and its initial speed is 14 m/s.
Mass of ball C is 0.7 kg and its initial speed is 2 m/s.
We need to find the final speed of the balls 1 s after being thrown. They all are thrown upward under the action of gravity. The equation of motion is : v = u -gt, g = 10 m/s²
For ball A,
v = 8-10(1) = = -2 m/s (downward)
For B,
v = 14-10(1) = 4 m/s (upward)
For C,
v = 2-10(1) = -8 m/s (downward)
It means the ranking is B>A>C i.e. the speed of ball B is the most and that of C is least.
All the balls are moving under the action of gravity. It would mean that the acceleration for all balls is same i.e. 10 m/s²
Net force is basically the force an object has when changing direction, so the answer would be D.
The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.
<u>Given the following data:</u>
- Acceleration = 2

To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:
First of all, we would calculate the total mass of the cat/elevator pair.

Total mass = 99 kilograms
Mathematically, Newton's Second Law of Motion is given by this formula;

Substituting the given parameters into the formula, we have;

Net force = 198 Newton
Read more here: brainly.com/question/24029674
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr