Since the system itself is giving off heat, this is a
reduction in the internal energy.
heat = - 25,000 J
Since work is being done on the system, therefore it is
an additional energy to the system. Work is given as:
work = - P dV
work = - 1.50 atm (6 L – 12 L)
work = 9 L atm
Since it is given that 1 L atm is equivalent to 101.3 J,
therefore the total energy added is:
energy due to work = 9 L atm (101.3 J / 1 L atm)
energy due to work = 911.7 J
Therefore the total change in internal energy is the sum
of heat and energy due to work:
Change in internal energy = - 25,000 J + 911.7 J
Change in internal energy = - 24,088.3 J
<span>Therefore, approximately 24.1 kJ of energy is lost by the
system in the total process.</span>
<span>
</span>
<span>Answer:</span>
<span>-24.1 kJ</span>
396 years, 198 years out, and 198 more years to return in 2395, assuming an immediate response. Radio travels at the speed of light. A light-year is the distance light (or radio) travels in a standard earth year.
<span>I wouldn't hold your breath...
If you need more help I will be glad to help!:D
*~"AB84"~*
</span>
Let D be the total distance (say in meters) traveled by the train and T the time (say in seconds) it takes to do so. (Assume the train moves in a straight line in only one direction.) Then the average velocity of the train as it covers this distance is
v (ave) = D/T
We're told the train can traverse a distance of D/4 in a matter of T/2 seconds if it moves at a speed of 5 m/s. This means
D/4 = (5 m/s) (T/2)
⇒ 5 m/s = 1/2 D/T
⇒ v (ave) = D/T = 10 m/s
The potential energy of a body is the maximum at the highest position of that body. So for example, if the maximum position of the ball is at 3 m, there will be the maximum potential energy.
To get the half of the maximum potential energy, the ball should be at the half of the maximum position, which is 1.5 m.
Answer:
μ₁ = 0.1048
μ₂ = 0.1375
Explanation:
Using static equation can find in both point the moment and the forces so:
∑ M = F *d , ∑ F = 0
∑ M A = 0
N₁ * 3 - 200 * 9.81 * 1.5 = 0
N₁ = 981
∑ M y = 0
N₂ + 300 * ³/₅ - 981 - 20 * 9.81 = 0
N₂ = 997.2 N
∑ M C = 0
F₁ * 1.75 - 300 * ⁴/₅ * 0.75 = 0
F₁ = 102.86
∑ M B = 0
300 * ⁴/₅ * 1 - F₂ * 1.75 = 0
F₂ = 137.14 N
The Force F1 and F2 related the coefficients of static friction
F₁ = μ₁ * N₁ ⇒ 102.86 N = μ₁ * 981 ⇒ μ₁ = 0.1048
F₂= μ₂ * N₂ ⇒ 137.14 N = μ₂ * 997 ⇒ μ₂ = 0.1375