Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
They were going at a velocity 4.07m/s
<u>Explanation:</u>
Distance s =5 m
initial velocity u= 0.8 m/s
Acceleration a =1.6m/s2
We have to calculate the velocity with which they were going afterwards i.e final velocity.
Use the equation of motion
They were going with a velocity 4.07 m/s afterwards.
Answer:
a) t = 0.528 s
b) D = 1.62 m
Explanation:
given,
speed of the baseball = 3.75 m/s
angle made with the horizontal = 35°
height of the roof edge = 2.5 m
using equation of motion
4.9 t² + 2.15 t - 2.5 = 0
on solving the above equation
t = 0.528 s
b) D = v cos θ × t
D = 3.75 × cos 35° ×0.528
D = 1.62 m
Answer:
A stone that is dropped down into an empty well
Answer:
29.2 ft/s
Explanation:
The distance of the light's projection on the wall
y = 13 tan θ
where θ is the light's angle from perpendicular to the wall.
The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,
Angular speed = w = (2π/3)
w = (θ/t)
θ = wt = (2πt/3)
(dθ/dt) = (2π/3)
y = 13 tan θ
(dy/dt) = 13 sec² θ (dθ/dt)
(dy/dt) = 13 sec² θ (2π/3)
(dy/dt) = (26π/3) sec² θ
when θ = 15°
(dy/dt) = (26π/3) sec² (15°)
(dy/dt) = 29.2 ft/s
