Answer:
1.06 m
Explanation:
Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C
r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m = 1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m
The distance between them is 1.06 m
Answer:
Centrifugal force
Explanation:
The reacting force that is equal to and opposite in the direction to the centripetal force and tends to fling air out of the center of rotation of High and Low pressure systems is called centrifugal force.
Centrifugal force is force that causes an object moving in a circular path to move out and away from the center of it's path, it always centripetal force and the force is imaginary, which can only be felt and not seen.
Answer:
bottle might burst
Explanation:
because liquid inside will expand and exert pressure on walls of the bottle.
Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
Answer:
a) t = 0.74s
b) D = 4.76m
c) Vf = 5.35m/s
Explanation:
The ball starts rolling when Vf = ωf*R.
We know that:
Vf = Vo - a*t
ωf = ωo + α*t
With a sum of forces on the ball:
With a sum of torque on the ball:
Replacing both accelerations:
t=0.74s
The distance will be:
Final velocity:
Vf=5.35m/s
