Answer:
Explanation:
Expression for relative velocity
=
= (.54 + .82 )c/
= 1.36 c / 1.4428
= .94 c
β = .94
What is the acceleration of a 55 kg block of cement when pulled sideways with a net force of 356 N? Answer in units of m/s2.
1 answer:
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Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
Answer:
Vc = 2.41 v
Explanation:
voltage (v) = 16 v
find the voltage between the ends of the copper rods .
applying the voltage divider theorem
Vc = V x ()
where
- Rc = resistance of copper =
(l = length , a = area, ρ = resistivity of copper)
- Ri = resistance of iron =
(l = length , a = area, ρ₀ = resistivity of copper)
Vc = V x ()
Vc = V x ()
Vc = V x ()
where
- ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
- ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter
Vc = 16 x ()
Vc = 2.41 v