Answer:
636.4 J
Explanation:
The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)
Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r
U' = 4kq²/r
= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m
= 900 Nm²/√2 m
= 636.4 J
Answer:
the magnitude of the electric field is 1.25 N/C
Explanation:
The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s
ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V
Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube
ε = ∫E.ds
ε = Eds since E is always parallel to the side of the cube
= E∫ds ∫ds = 4L since we have 4 sides
= E(4L)
= 4EL
So,4EL = 0.05 V
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
So, the magnitude of the electric field is 1.25 N/C
<h2>
Answer:</h2>
Answer to this question is (A)
<h2>
Explanation</h2>
A ball bouncing on the floor is not the example of simple harmonic motion. SHM is the special kind of to and fro motion in which a particle oscillate about its mean position in a straight line. The acceleration of the particle is always directed towards its mean position and is directly proportional to its displacement from its mean position.
In case of a ball bouncing on the ground, the motion of the ball is not SHM, as neither it’s a to and fro motion nor the acceleration is proportional to its displacement from its mean position.
Answer: The planet, named after the Roman goddess of art and beauty, can actually be bright enough to cast shadows on a moonless night. It appears so close to the sun because its orbital radius is smaller than the Earth's, and because it also moves faster than Earth, its orbital period is shorter.
Answer:
the pressure exerted on the turf when the player lands on his knee is 240,000 N/m².
Explanation:
Given;
combined weight of the players, F = 2400 N
measurement of the player's knee = 0.1 m by 0.1 m
The area of the player's knee is calculated as;
A = 0.1 m x 0.1 m = 0.01 m²
The pressure exerted on the turf when the player lands on his knee is calculated as;
P = F / A
P = (2400 N) / (0.01 m²)
P = 240,000 N/m²
Therefore, the pressure exerted on the turf when the player lands on his knee is 240,000 N/m².
