Answer:
.
Explanation:
The average kinetic energy per molecule of a ideal gas is given by:

Now, we know that 
Before the absorption we have:
(1)
After the absorption,
(2)
If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)




Therefore the ratio will be 
I hope it helps you!
Answer:
velocity = distance / time taken
= 200/4
= 50 m/s
is the correct answer
Answer:
Explanation:
Given that,
Mass per unit length is
μ = 4.87g/cm
μ=4.87g/cm × 1kg/1000g × 100cm/m
μ = 0.487kg/m
Tension
τ = 16.7N
Amplitude
A = 0.101mm
Frequency
f = 71 Hz
The wave is traveling in the negative direction
Given the wave form
y(x,t) = ym• Sin(kx + ωt)
A. Find ym?
ym is the amplitude of the waveform and it is given as
ym = A = 0.101mm
ym = 0.101mm
B. Find k?
k is the wavenumber and it can be determined using
k = 2π / λ
Then, we need to calculate the wavelength λ using
V = fλ
Then, λ = V/f
We have the frequency but we don't have the velocity, then we need to calculate the velocity using
v = √(τ/μ)
v = √(16.7/0.487)
v = 34.29
v = 5.86 m/s
Then, we can know the wavelength
λ = V/f = 5.86 / 71
λ = 0.0825 m
So, we can know the wavenumber
k = 2π/λ
k = 2π / 0.0825
k = 76.18 rad/m
C. Find ω?
This is the angular frequency and it can be determined using
ω = 2πf
ω = 2π × 71
ω = +446.11 rad/s
D. The angular frequency is positive (+) because the direction of propagation of wave is in the negative direction of x
Answer:
72%
Calculating Efficiency:
The amount of work obtained from a machine is always less than the amount of work put into it. Thisis because some of the work is lost due to friction. The efficiency of a machine can be calculated usingthe following formula:Percent Efficiency = (work output-Machine / work input-You) x 100%
Answer:
FC vector representation

Magnitude of FC

Vector direction FC
degrees: angle that forms FC with the horizontal
Explanation:
Conceptual analysis
Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.
The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.
degrees
To calculate the magnitudes of the forces we apply Coulomb's law:
Equation (1): Magnitude of the electric force of the charge qA over the charge qC
Equation (2)
: Magnitude of the electric force of the charge qB over the charge qC
Known data





Problem development
In the equations (1) and (2) to calculate FAC Y FBC:


Components of the FBC force at x and y:


Components of the resulting force acting on qC:


FC vector representation

Magnitude of FC

Vector direction FC
degrees: angle that forms FC with the horizontal