Answer:
Time, t = 80 seconds
Explanation:
Given that,
The frequency of the oscillating mass, f = 1.25 Hz
Number of oscillations, n = 100
We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :



t = 80 seconds
So, it will make 100 oscillations in 80 seconds. Hence, this is the required solution.
Fruits and frozen fruit bars is the correct answer.
Answer:
option (B)
Explanation:
Intensity of unpolarised light, I = 25 W/m^2
When it passes from first polarisr, the intensity of light becomes

Let the intensity of light as it passes from second polariser is I''.
According to the law of Malus

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 11.66 W/m^2
I'' = 11.7 W/m^2
Answer:
(a) 0.2618 J
(b) 0.1558 J
(c) 0 J
Explanation:
from Hook's Law,
The energy stored in a stretched spring = 1/2ke²
Ep = 1/2ke² ......................... Equation 1
Where k = spring constant, e = extension, E p = potential energy stored in the spring.
(a) When The spring is stretched to 4.11 cm,
Given: k = 310 N/m, e = 4.11 cm = 0.0411 m
Substituting these values into equation 1
Ep = 1/2(310)(0.0411)²
Ep = 155(0.0016892)
Ep =155×0.0016892
Ep = 0.2618 J.
(b) When the spring is stretched 3.17 cm
e = 3.17 cm = 0.0317 m.
Ep = 1/2(310)(0.0317)²
Ep = 155(0.0317)²
Ep = 155(0.0010049)
Ep = 0.155758 J
Ep ≈ 0.1558 J.
(c) When the spring is unstretched,
e = 0 m, k = 310 N/m
Ep = 1/2(310)(0)²
Ep = 0 J.