Answer:
Diana's speed relative to ground is <u>16 km/h</u> in the direction of motion of train.
Explanation:
Given:
Velocity of train in forward direction is,
Here,
Velocity of Diana relative to train in the backward direction is,
Negative sign implies backward motion or motion opposite to the direction of train's motion. Here, Diana is walking from front of train to back. So, Diana is moving in the opposite direction.
Now, we know that, for two bodies 'A' and 'B', velocity of 'A' relative to ground is given as:
Therefore, velocity of Diana relative to ground is given as:
So, Diana's speed relative to ground is 16 km/h in the direction of motion of train.
Hydroelectric dams convert the gravitational potential energy of falling water into electricity.
Answer:
(a) Surface charge density is the charge per unit area.
(b)
(c)
Explanation:
(a) Surface charge density is the charge per unit area. The area of the square plate can be calculated by its side length.
Half of the total charge is distributed on one side and the other is distributed on the other side.
Therefore, surface charge density on each face of the plate is
(b) To find the electric field just above the plate, Gauss' Law can be used. Normally, Gauss' Law can only be used in infinite sheet (considering the flat surfaces), but just above the surface can be considered that the distance from the surface is much much smaller than the length of the plate (x << l).
In order to apply Gauss' Law, we have to draw an imaginary cylinder with radius r. The cylinder has to stay perpendicular to the plane.
The direction of the electric field is in the upwards direction.
(c) The magnitude of the electric field is the same as that of upper side. Only the direction is reversed, downward direction.
Answer:
v = 4t - t²/3
Explanation:
4 - k*6 = 0
k = 4/6 = 2/3
dv/dt = 4- 2/3 t
integral of dv = integral of (4-2/3 t) dt
v = 4t - t²/3
Answer:
(a) (18m/s/t₁)m/s²
(b) -12.5m/s²
(c) -20mls²
Explanation:
(a) Let t₁ be the initial time
a = v-u/t
acc = (18m/s/t₁)m/s²
(b) acc = -30m/s/2.4
= -12.5m/s²
(c)The particle was at a speed of 18m/s in the positive x-direction and later after 2.4s ≡Δt, it was at speed of -30m/s in the negative x-direction.
so this imply that the velocity was first v₁ =18m/s and later v₂ = -30m/s.
The average acceleration is then:
Aavg =<u> Δv</u>
Δt
= v₂-v₁/Δt
= -30-18/2.4 = -20mls²