Answer: Ax=(Vx-Vox)/(T)
Vx=Vox+Ax*T
Solving for Ax in terms of Vx, Vox, T
Vx-Vox=Ax*t
Ax=(Vx-Vox)/(T)
This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.
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Internal energy of the system changes by ΔE = 178 J.
Heat given to the system = Q = +658 J.
According to the first law of thermodynamics,
ΔE = Q + W
178 = 658 + W
∴ W = 178-658 = -480 J
Minus sign indicates that work is done by the system.
What is the heat extracted from the cold reservoir for the refrigerator shown in(Figure 1) ? Assume that W1 = -123J and W2 = 88J .
<span>Qc= _________ </span>
<span>Part B
</span>
K=105J
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
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
