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3 years ago

Identical rays of light in air are refracted upon entering three transparent materials:

Physics

1 answer:

aev [14]3 years ago

3 0

Answer:

Explanation:

for us to rank them we need to calculate the angle of refraction for the various materials. The angle of refraction is the angle between the refracted ray and the normal.

refractive index(n) = velocity of light in air/ velocity of light in medium

refractive index (n) = sini/sinr

lets equate the right side of both equations.

$\frac{sini}{sinr}=\frac{c}{velocity of light in medium}$

we will assume 30 degree as the angle of incidence for all materials.

(a) water

$\frac{sin30}{sinr}=\frac{c}{0.75c}$

sinr = 0.75c/c *sin30

sinr = 0.375

r= arcsin0.375

r=22.02°

(b) ethyl alcohol

$\frac{sin30}{sinr}=\frac{c}{0.7c}$

sinr = 0.7c/c *sin30

sinr = 0.35

r= arcsin0.35

r=20.49°

$\frac{sin30}{sinr}=\frac{c}{0.6c}$

sinr = 0.6c/c *sin30

sinr = 0.3

r= arcsin0.35

r=17.46°

it is observed that the angle of refraction decreases as the velocity of light in glass decreases.

ranking the materials according to how much the light ray bends toward the normal, from most bending to least bending we have;

Crown glass > ethyl alcohol> water

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Which of the following are examples of centripetal acceleration? Check all that apply.

k0ka [10]

<u>Answer:</u> The correct answer is option B, C and E.

<u>Explanation:</u>

Centripetal acceleration is defined as the acceleration win which an object moves in a curved path. Formula for this acceleration is given by the equation:

$a_c=\frac{v^2}{r}$

where,

$a_c$ = centripetal acceleration

v = linear speed of the object

r = radius of the curved path

From the given options,

Option A: As, the golf ball is not moving in a curved path. Hence, it is not an example of centripetal acceleration.

Option B: As, a car is moving in a curved path. Hence, it is an example of centripetal acceleration.

Option C: As, a person is moving in a curved path. Hence, it is an example of centripetal acceleration.

Option D: As, a car is not moving in a curved path and is moving in a straight road. Hence, it is not an example of centripetal acceleration. The car is moving with zero acceleration because the direction of the car is not changing.

Option E: As, a bicyclist is moving in a curved path which is around the lake. Hence, it is an example of centripetal acceleration.

6 0

3 years ago

A flywheel in the form of a uniformly thick disk of radius 1.33 m1.33 m has a mass of 70.6 kg70.6 kg and spins counterclockwise

ladessa [460]

Answer:

The constant torque required to stop the disk is 8.6 N-m in clockwise direction .

Explanation:

Let counterclockwise be positive direction and clockwise be negative direction .

Given

Radius of disk , r = 1.33 m

Mass of disc , m = 70.6 kg

Initial angular velocity , $\omega_i =217 rpm$

Final angular velocity , $\omega_f =0\, rpm$

Time taken to stop , t = 2.75 min

Let $\alpha$ be the angular acceleration

We know

$\omega _f=\omega _i+\alpha t$

=> $0=217+2.75\alpha =>\alpha = -78.9\frac{rev}{min^{2}}$

=> $\alpha =-\frac{78.9\times 2\pi}{60\times 60}\frac{rad}{s^{2}}=-0.138 \frac{rad}{s^{2}}$

Torque required to stop is given by

$\tau =I\alpha$

where moment of inertia , $I=\frac{mr^{2}}{2}=\frac{70.6\times 1.33^{2}}{2}kg.m^{2}=62.5 kg.m^{2}$

=> $\therefore \tau =-0.138\times 62.5\, N.m=-8.6\, N.m$

Thus the constant torque required to stop the disk is 8.6 N-m in clockwise direction .

3 0

2 years ago

Consider a keen little boy who is having a wagon race with a friend. He starts from rest and

MAXImum [283]

A) His wagon will accelerate more.
B) His wagon will accelerate less. Both parts are answered by F=ma. Mass is inversely proportional to acceleration, and force is directly proportional to acceleration.

7 0

2 years ago

Unit of gravitational force

azamat

Answer:

Newton

Explanation:

9.80665 × kgf = 1 Newton N

1 kilogram force (kgf) was the force of gravity, which is pushing a mass of 1 kg at one place in the world on the ground; calculated according to Newton's law force = mass × acceleration.

3 0

3 years ago

ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly

Igoryamba

The period T of a pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$ . Using this data, we can solve the previous formula to find L:
$L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm$

4 0

2 years ago