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Likurg_2 [28]
3 years ago
8

An incident ray strikes a boundary at a 72.5-degree angle, resulting in an angle of refraction of 39.6 degrees. The index of ref

raction of the first medium is 1.0003.
What is medium 1?



What is medium 2?
Physics
2 answers:
maks197457 [2]3 years ago
7 0
A. Medium 1 is air. The refractive index of air is 1.0003. Refractive index is a measure of the bending of a ray of light when passing from one medium to another. If i is the angle of incidence of a ray in vacuum and r is the angle of refraction, the refractive index n is defined as the ratio of the sine angle of incidence to the sine of the angle of refraction; that is, n= sin i/ sin r. Refractive index may also be given by the velocity of light in medium 1 divided by the velocity of light in medium 2.


b. In this case the refractive index = sin 72.5/sin 39.6 
                                                   = 1.4962 
but n = x/ 1.0003 = 1.4962
Therefore; x = 1.4967
Hence the refractive index of medium 2 is 1.4967. 
I therefore think that medium 2 is toluene; since it has a similar refractive index.

Aloiza [94]3 years ago
4 0

Medium 1: Air

Medium 2: Lucite

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inessss [21]

Answer:

The term soil fabric refers to the geometrical arrangement of individual particles in a soil, including the geometrical distribution of pore spaces. Soil fabric may be used to describe the particle arrangement in both cohesive soils such as clays and granular soils such as silts, sands, and gravels.

5 0
2 years ago
A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read
timama [110]

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

7 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

5 0
3 years ago
Please help me with my Physical Science! 50 POINTS
DaniilM [7]

Answer:

1, When Jane brakes, the brakes slow the car wheels turning and the road surface exerts a backwards force on the tires, causing the car to decelerate. The pocket book tends to continue on in a straight line (Newton's first law). If she brakes hard enough that the friction between the book and the car seat is insufficient to decelerate the book as fast as the car is decelerating, the book will slide off the seat, and gravity pulls it to the floor

2.

When the diver uses his / her force to depress the springboard, the springboard pushes him back with equal force

3.Newton's Second Law (F=ma)

4. 5 N

5. 19.5 N

65kg * 0.3 m/s^2

6.0.2 N/s

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7.-Walking then pushing the moving forward

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Explanation:

6 0
3 years ago
Read 2 more answers
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