Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
1. Giga is the largest
2. Stem-and-leaf
Answer:
A) 140 k
b ) 5.22 *10^3 J
c) 2910 Pa
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J
number of moles (n) = 3
A ) calculate the change in temp of the gas
since the volume of gas is constant no work is said to be done
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R
= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
= 140 K
B) Calculate the change in its internal energy
ΔU = Q this is because no work is done
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure
applying ideal gas equation
P = nRT/V
therefore ; Δ P = ( n*R*ΔT/V )
= ( 3 * 8.3144 * 140 ) / 1.20
= 2910 Pa
Complete question:
In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)
Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m
What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?
Answer:
The rotational speed of the ring module have to be 0.92 rad/s
Explanation:
Given;
the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m
When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.
Centripetal acceleration, a = g = 9.8 m/s²
Centripetal acceleration, a = v²/r
But v = ωr
a = g = ω²r

Therefore, the rotational speed of the ring module have to be 0.92 rad/s
Answer:
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