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Norma-Jean [14]
3 years ago
6

A light ray is incident from air into glass (ng = 1.52) then onto water (nw= 1.33). The wavelength of light in air (na= 1) is ai

r = 500 nm and it travels at a speed c = 3 x 108 m/s. The wavelength of light, A, and its frequency, f, in water, are, respectively:
(a) 376 nm, 6 x 10¹4 Hz.
(b) 376 nm, 8 x 10¹¹ Hz
(c) 500 nm, 6 x 10¹4 Hz
(d) 500 nm, 8 x 10¹¹ Hz​
Physics
1 answer:
gregori [183]3 years ago
3 0
The answer is is b hope this helped
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Visible light falls into wavelength ranges of 400-700 nm, for which 1 m = 1 × 10 9 nm . The energy and wavelength of light are r
Paul [167]

Answer:

E = 3.54 x 10⁻¹⁹ J

Explanation:

The energy of the photon can be given in terms of its wavelength by the use of the following formula:

E = \frac{hc}{\lambda}

where,

E = energy = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light = 2.998 x 10⁸ m/s

λ =  wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m

Therefore,

E = \frac{(6.626\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{5.606\ x\ 10^{-7}\ m}\\

<u>E = 3.54 x 10⁻¹⁹ J</u>

7 0
3 years ago
Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

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tom and ted are sitting on seprate chairs that have wheels. tom pushes ted and, in turn, he starts moving too. in which directio
IceJOKER [234]
In the direction opposite to his push.
5 0
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a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
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