Answer:
one with 60g have the more potential
Explanation:
Answer:
Part a)

Part b)

Explanation:
As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved
so we can say




Part b)
magnitude of the initial speed of A = 
magnitude of the initial speed of B = 
magnitude of final speed of A = 
magnitude of final speed of B = 
Now change in total kinetic energy is given as



The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.
for 1 degree................... 7 Joules
y given degree........ p Joules
p=7y
In our case y=(t-25) .
h(t) = 7(t-25) which is the final answer.
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