Use the eq. of Young modulus Y=(F/A)/(∆l/lo)
dimana ∆l is the elongation of wire, lo is its initial length.
So ∆l = (F/A)lo/Y.
∆l = (1000N/(6.5 × 10^-7 m^2))×(2.5m)/(2.0 × 10^-11 N/m^2)
Use calculator to finish it.
Answer:
100 newtons
Explanation:
Given,
Jamal pushing a large box by a force, F = 100 N
Work done on the large box is, W = 0
It is because the applied force is less than the force of the friction between the two surfaces.
Yet, there will be a force that is exerted by the large box on Jamal.
According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

Hence, If the action force is 100 N, then the reaction force should be in 100 N
I this is tricky one sec.. I would go with b
Answer:
0.063 Kg
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 10 Hz
Spring constant (K) = 250 N/m
Mass (m) =?
Next, we shall determine the period of oscillation. This can be obtained as follow:
Frequency (f) = 10 Hz
Period (T) =?
T = 1/f
T = 1/10
T = 0.1 s
Finally, we shall determine the mass of the spring. This can be obtained as follow:
Spring constant (K) = 250 N/m
Period (T) = 0.1 s
Pi (π) = 3.14
Mass (m) =?
T = 2π√(m/K)
0.1 = 2 × 3.14 × √(m/250)
0.1 = 6.28 × √(m/250)
Divide both side by 6.28
0.1 / 6.28 = √(m/250)
Take the square of both side.
(0.1 / 6.28)² = m/250
Cross multiply
m = (0.1 / 6.28)² × 250
m = 0.063 Kg
Therefore, the mass of the spring is 0.063 Kg.