Answer:
Explanation:
Expression for relative velocity
= 
= (.54 + .82 )c/ 
= 1.36 c / 1.4428
= .94 c
β = .94
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:
Both speed and velocity are changing.
Explanation:
They are both going up so both are changing
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
Ohms Law: V = IR
V is the voltage in volts
I is the current in amps
R is the resistance in Ohms
Rearrange: R = V/I
R = (110)/(0.050)
R = 2200
There are 2200 Ohms of resistance in the circuit.