<h2>Explanation:</h2><h3>3. </h3>
When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:
- Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.
_____
<h3>4.</h3>
The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...
n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33
- none of the offered choices is correct. The closest is 1.34.
Answer:
The mass of the block is 1250g.
Explanation:
Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :

Let density = 250,
Let volume = 5,



In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:
E=hv
where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s
Answer:
distance is 13 m for 100 dB
distance is 409 km for 10 dB
Explanation:
Given data
distance r = 2.30 m
source β = 115 dB
to find out
distance at sound level 100 dB and 10 dB
solution
first we calculate here power and intensity and with this power and intensity we will find distance
we know sound level β = 10 log(I/
) ......................a
put here value (I/
) = 10^−12 W/m² and β = 115
115 = 10 log(I/10^−12)
so
I = 0.316228 W/m²
and we know power = intensity × 4π r² ...............b
power = 0.316228 × 4π (2.30)²
power = 21.021604 W
we know at 100 dB intensity is 0.01 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 0.01 × 4π r²
so by solving r
r = 12.933855 m = 13 m
distance is 13 m
and
at 10 dB intensity is 1 × 10^–11 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 1 × 10^–11 × 4π r²
by solving r we get
r = 409004.412465 m = 409 km