Elastic potential energy stored in a spring is
(1/2) · (spring constant) · (stretch or compress)² .
PE = (1/2) · (100 N/m) · (0.1 m)²
PE = (50 N/m) · (0.01 m²)
PE = (50 · 0.01) (N · m / m²)
PE = 0.5 N · m
PE = 0.5 Joule
Answer:
The speed of waves on this wire is 329.14 m/s
Explanation:
Given;
tension of the wire, T = 650 N
mass per unit length, μ = 0.06 g /cm = 0.006 kg/m
(convert the unit of mass per length in g/cm to kg/m by dividing by 10 = 0.06 / 10 = 0.006 kg/m)
The speed of waves on this wire is given as;
Therefore, the speed of waves on this wire is 329.14 m/s
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ
= m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s
<span>Electrical discharge from a charged object
is your answer</span>
It produces only virtual images is the answer