All matter is made up of atoms
Answer:
ΔU = - 310.6 J (negative sign indicates decrease in internal energy)
W = 810.6 J
Explanation:
a.
Using first law of thermodynamics:
Q = ΔU + W
where,
Q = Heat Absorbed = 500 J
ΔU = Change in Internal Energy of Gas = ?
W = Work Done = PΔV =
P = Pressure = 2 atm = 202650 Pa
ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³
Therefore,
Q = ΔU + PΔV
500 J = ΔU + (202650 Pa)(0.004 m³)
ΔU = 500 J - 810.6 J
<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>
<u></u>
b.
The work done can be simply calculated as:
W = PΔV
W = (202650 Pa)(0.004 m³)
<u>W = 810.6 J</u>
Answer:
What are we supposed to find, if it is kinetic energy then this is the solution.
K.E=1/2mv^2
K.E= kinetic energy
M=mass
V=velocity
K.E =0.5*55*0.6^2
K.E=9.9J
Explanation:
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) =
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ =
–K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J
Answer:
b. they get blown in from colder or warmer areas.