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
Answer:
Travelled 18 km, they are 6 km from home.
Explanation:
12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.
Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.
Energy in a machine can be something technical like a wire or something went wrong with the system