The frequency of the light source is 1.5 x 10¹⁵ Hz.
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Frequency of the light source</h3>
The frequency of the light source is determined using the following equations;
c = fλ
where;
c is speed of light
f is the frequency
λ is the wavelength
f = (3 x 10⁸) / (2 x 10⁻⁷)
f = 1.5 x 10¹⁵ Hz
Thus, the frequency of the light source is 1.5 x 10¹⁵ Hz.
Learn more about frequency of light here: brainly.com/question/10728818
Answer:
b) true. The jobs are equal
Explanation:
The work on a body is the scalar product of the force applied by the distance traveled.
W = F. d
Work is a scalar, the work equation can be developed
W = F d cos θ
Where θ is the angle between force and displacement
Let's apply these conditions to the exercise
a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same
b) true. The jobs are equal dx = d cosθ
c) False, because the force is equal and the projection of displacement is the same
d) False, knowledge of T is not necessary because the projection of displacement is always the same
e) False mass is not in the definition of work
Answer:
3 m/s^2
Explanation:
acceleration= Change in velocity/time
= 30-0 / 10
= 30/10
=3 m/s^2
Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J