Answer: Before the jump, the snowboarder would carry potential energy.
During the jump he will carry kinetic energy.
And after the jump, assuming hes at a full stop, he will carry potential energy once again.
The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.
Explanation:
https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/
Answer:
Isabella will not be able to spray Ferdinand.
Explanation:
We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =.?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.9
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s
Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:
Horizontal velocity (u) = 3.5 m/s
Time (t) = 0.45 s
Horizontal distance (s) =?
s = ut
s = 3.5 × 0.45
s = 1.58 m
Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:
Ferdinand's position = 10 m
Distance travelled by the water = 1.58 m
From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.
Answer:
∆T = Mv^2Y/2Cp
Explanation:
Formula for Kinetic energy of the vessel = 1/2mv^2
Increase in internal energy Δu = nCVΔT
where n is the number of moles of the gas in vessel.
When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas
We say
1/2mv^2 = ∆u
1/2mv^2 = nCv∆T
Since n = m/M
1/2mv^2 = mCv∆T/M
Making ∆T subject of the formula we have
∆T = Mv^2/2Cv
Multiple the RHS by Cp/Cp
∆T = Mv^2/2Cv *Cp/Cp
Since Y = Cp/CV
∆T = Mv^2Y/2Cp k
Since CV = R/Y - 1
We could also have
∆T = Mv^2(Y - 1)/2R k
