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.
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
Answer:
The mass number 204 – 82 protons = 122 neutrons
Explanation:
Hope this helps!
Answer:
61.33 Kg
Explanation:
From the question given above, the following data were obtained:
Distance = 1×10² m
Time = 9.5 s
Kinetic energy (KE) = 3.40×10³ J
Mass (m) =?
Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:
Distance = 1×10² m
Time = 9.5 s
Velocity =?
Velocity = Distance / time
Velocity = 1×10² / 9.5
Velocity = 10.53 m/s
Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:
Kinetic energy (KE) = 3.40×10³ J
Velocity (v) = 10.53 m/s
Mass (m) =?
KE = ½mv²
3.40×10³ = ½ × m × 10.53²
3.40×10³ = ½ × m × 110.8809
3.40×10³ = m × 55.44045
Divide both side by 55.44045
m = 3.40×10³ / 55.44045
m = 61.33 Kg
Thus, the mass of Leroy Burrell is 61.33 Kg
