The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
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Explanation:
It is given that,
Uncertainty in the speed of an electron, 
According to Heisenberg uncertainty principle,
is the uncertainty in the position of an electron
Since, 
So, the uncertainty in its position is 
. Hence, this is the required solution.
Answer:
33.6 Ns backward.
Explanation:
Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.
From Newton's second law of motion,
Impulse = change in momentum
I = mΔv................................. Equation 1
Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.
Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)
Substituting into equation 1
I = 28(-1.2)
I = -33.6 Ns
Thus the impulse = 33.6 Ns backward.
Answer:
13.78 mT
Explanation:
The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,
A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength
So,
B = ε/ωNA
substituting the values of the variables into the equation given that ε = 17 V
So, B = ε/ωNA
B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)
B = 17 V/(1233.8634 rad-turns-m²/s)
B = 0.01378 T
B = 13.78 mT
