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
Answer:
Wave speed = Wavelength x Frequency
26 m/s = Wavelength x 49
Divide by 49 to find the wavelength:
The wavelength is approximately 0.53
Let me know if this helps!
Answer:
<em>B. 68.6m</em>
Explanation:
<u>Free Fall Motion
</u>
When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is
.

We know the egg takes 3.74 seconds to reach the ground. The height it was launched from is


The closest correct option is
B. 68.6m
The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The specific heat capacity of a substance can be calculated using the following formula:
Q = m × c × ∆T
Where;
- Q = quantity of heat absorbed (J)
- c = specific heat capacity (4.18 J/g°C)
- m = mass of substance
- ∆T = change in temperature (°C)
According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:
2510 = 158 × c × (61°C - 32°C)
2510 = 4582c
c = 2510 ÷ 4582
c = 0.5478 J/g°C
Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.
Learn more about specific heat capacity at: brainly.com/question/2530523
Answer:

Explanation:
By Einstein's Equation of photoelectric effect we know that

here we know that
= energy of the photons incident on the metal
= minimum energy required to remove photons from metal
= kinetic energy of the electrons ejected out of the plate
now we know that it requires 351 nm wavelength of photons to just eject out the electrons
so we can say

here we know that

now we have

now by energy equation above when photon of 303 nm incident on the surface




