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1 year ago

Which is the BEST example of refraction?

Physics

2 answers:

Mademuasel [1]1 year ago

8 0

Answer:

Option (c) is correct.

Explanation:

When a ray of light passes from one optical medium to another optical medium, the path of ray of light is deviated, this phenomena is called refraction.

It occurs due to the change in speed of light, as it passes from medium to another.

So, when a laser beam passes through a glass block, the refraction takes place.

Aneli [31]1 year ago

6 0

Answer:

c would be your answer just did this on usatest prer

Explanation:

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A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an

Cloud [144]

Answer:

1.36

Explanation:

$n_{air}$ = Index of refraction of air = 1

$n_{plastic}$ = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

$n_{air}$ Sini = $n_{plastic}$ Sinr

(1) SIn32.0 = $n_{plastic}$ Sin23.0

$n_{plastic}$ = 1.36

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2 years ago

1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour

grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 $kg \cdot cm^2$

Explanation:

Given:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

To Find:

The moment of inertia of Alex's rolling hoop = ?

Solution:

The moment of inertia = $mr^2$

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia = $(0.350)(0.75)^2$

=> moment of inertia = $(0.350)(0.5625)$

=> moment of inertia = $(0.197)$

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 $kg m^2 s^{-1}$

If she pulls her arms in to 0.12 m, her new linear speed is $18.33 m/s^2$

Explanation:

Given:

Mass = 2.0 kg

Radius = 0.8 m

Velocity = 1.2 m/s

a.The combined angular momentum of the masses:

$L = r \cdot m \cdot v_1$

Substituting the values,

$L = 0.8 \cdot 2.0 \cdot 1.1$

L= 1.76 $kg m^2 s^{-1}$

b. If she pulls her arms in to 0.12 m, what is her new linear speed

$0.12 \cdot 0.8 \cdot v_2 = 1.76$

$0.096 cdot v_2 = 1.76$

$v_2 = \frac{1.76}{0.096}$

$v_2 = 18.33 m/s^2$

6 0

1 year ago

A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

dybincka [34]

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

$PE = \dfrac{1}{2}kx^2$

$PE = \dfrac{1}{2}\times 5000\times 0.025^2$

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

5 0

1 year ago

A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

6 0

1 year ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

2 years ago