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ASHA 777 [7]
2 years ago
10

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

and nviolet = 1.60.

Physics
1 answer:
Harlamova29_29 [7]2 years ago
6 0
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


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Two objects of the same mass move along the same line in opposite directions. The first mass is moving with speed . The objects
Alexandra [31]

Explanation:

Let the mass of two objects be m. Both objects move along the same line in opposite directions. Let v and u₂ are speeds of both objects before collision.

After the collision, both objects stick together and move with the speed of 0.1 V the direction of the velocity of the first mass before the collision.

Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)v

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{m(u_1+u_2)}{(m_1+m_2)}

0.1v=\dfrac{m(v+u_2)}{(2m)}

On solving above equation, u_2=-0.8v

So, the speed of the second mass before the collision is 0.8 v. The negative sign shows that the it moves in opposite direction. Hence, this is the required solution.

6 0
3 years ago
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
3 years ago
A student discovers that sound waves travel 1,687.5 meters in 5 seconds through air at a temperature of 10°C. Based on this info
vitfil [10]

Answer:

The speed of sound, in m/s, through air at this temperature is 343.5 m/s

Explanation:

Given;

distance traveled by sound, d = 1,687.5 meters

time taken for the sound to travel, t = 5 seconds

air temperature, θ = 10°C

Speed of sound = distance traveled by sound / time taken for the sound to travel

Speed of sound = d / t

                           = 1687.5 m / 5 s

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Speed of sound at the given temperature is calculated as;

c = 337.5 + 0.6θ

c = 337.5 + 0.6 x 10

c = 337.5 + 6

c = 343.5 m/s

Therefore, the speed of sound, in m/s, through air at this temperature is 343.5 m/s

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Answer:

Explanation:

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