You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe
1 answer:
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Answer:
a
b
Explanation:
From the question we are told that
The wavelength of the light is
The distance of the slit separation is
Generally the condition for two slit interference is
Where m is the order which is given from the question as m = 2
=>
substituting values
Now on the second question
The distance of separation of the slit is
The intensity at the the angular position in part "a" is mathematically evaluated as
Where is mathematically evaluated as
substituting values
So the intensity is
The energy of a photon is given by:
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon,
, so we can rearrange the equation to calculate the frequency of the photon:
And now we can use the following relationship between frequency f, wavelength
and speed of light c to find the wavelength of the photon:
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon,
And now we can use the following relationship between frequency f, wavelength
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s