Answer:
12m
Explanation:
To obtain the answer to the question given, we must observe the characteristics of image formed by a plane mirror.
The image formed by a plane mirror have the following characteristics:
1. Laterally inverted.
2. Same distance as the object from the mirror.
3. Same height as the object.
4. Virtual.
With the above information, we can calculate the distance between the boy and his image as follow:
Initially:
Object distance (u) = 4m
Image distance (v) = 4m
The boy moved 2m away, therefore:
Object distance (u) = 2 + 4 = 6m
Image distanc(v) = 2 + 4 = 6m
The distance between the boy and his image will be the sum of his distance (u) and image distance (v) i.e (u + v)
The distance between the boy and his image = 6 + 6 = 12m
Therefore, the distance between the boy and his image is 12m.
Since it was stated that it must move at constant
velocity, so the only force it must overpower is the frictional force.
So the equation is:
F cos θ = Ff
F cos 36 = 65 N
F = 80.34 N
<span>So the nurse must exert 80.34 N of force</span>
Answer:

Explanation:
As we know that the position of maximum intensity on the screen is given as

here we know that
= wavelength
L = distance of the screen
d = distance between two slits
now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen
so we have

so here we have


Answer:
Conservation of angular momentum
Explanation:
When the objects spread in universe after big bang, because of the tremendous force , they gained angular momentum and started to rotate. Since, then the object continue to rotate on their axis because of conservation of angular momentum. In vacuum of space there no other forces that can stop these rotation, therefore, they continue to rotate.
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