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4 years ago

why doesn't light bend (refract) when it initially enters a glass? what actually happens to the light inside the semi-circle?

1 answer:

7 0

The light does not bend when it initially enters the glass because it travels from a less dense medium such as air to a dense medium such as glass. The refraction of light happens only on the flat side of the semi-circle, while on the circular side, it strikes on the interface.

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Jim began a 153-mile bicycle trip to build up stamina for a triathlon competition. Unfortunately, his bicycle chain broke, so he

NARA [144]

consider the motion when Jim travels by bicycle

D = total distance to be traveled for the trip = 153 mile

v = speed of riding = 42 mph

t = time taken to complete the trip by bicycle = ?

using the equation

t = D/ v

inserting the values

t = 153/42

t = 3.64 h

while walking and riding :

T = total time taken to complete the trip by walking and riding = 5 hours

t ' = time of walking = T - t = 5 - 3.64 = 1.36 hours

v' = speed of walking = 4 mph

d' = distance traveled by walking = ?

distance traveled by walking is given as

d' = v' t'

d' = 4 x 1.36

d' = 5.44 miles

d = distance traveled by bicycle = D - d' = 153 - 5.44 = 147.56 miles

v = speed of riding = 42 mph

t = time spent on the bicycle = ?

time spent on the bicycle is given as

t = d/v

t = 147.56/42

t = 3.51 hours

5 0

3 years ago

If you blow air between a pair of closely-spaced Ping-Pong balls suspended by strings, the balls will swing

sveta [45]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago

Which of the following would decrease the resistance to the flow of an electric current through a body?

Art [367]

Shortening the conductor (B) <span> would decrease the resistance
to the flow of an electric current through a body.</span>

8 0

3 years ago

The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation b

d1i1m1o1n [39]

The electrostatic force between two charges Q1 and q is given by
$F=k_e \frac{Q_1 q}{r^2}$
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges

Re-arranging the formula, we have
$q= \frac{F r^2}{k_e Q_1}$
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
$q= \frac{(0.961 N)(0.67 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.44\cdot 10^{-6}C)}=5.69 \cdot 10^{-6} C$

And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.

6 0

3 years ago