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REY [17]
3 years ago
5

A beam of visible light is passed through a plexiglass container filled with water. If a beam of light comes in at a slight angl

e, describe the change in the path of the beam of light as it passes through the container, moving in from one side and then out the other.
A) The beam of light will bend twice: entering and leaving the container.
B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
C) The beam of light will refract four times; largest entering container; smaller angles of refraction moving through water, through container wall again, and finally into the air on the outside.
D) The beam of light will bend three times: large angle of refraction through container wall; smaller angle of refraction through water; greater angle passing out of a container.
Physics
2 answers:
pentagon [3]3 years ago
8 0

B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
patriot [66]3 years ago
8 0

Answer:

C) The beam of light will refract four times; largest entering container; smaller angles of refraction moving through water, through container wall again, and finally into the air on the outside.

Explanation:

Just got it right on the test

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two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters
Free_Kalibri [48]

Answer:

0 m/s²

Explanation:

Since each skater pushes the other with a force of 70 N, according to Newton's third law, there is an equal reaction and thus the other pushes back with a force of 70 N in the other direction, so we have forces of +70N and -70 N respectively. So, the net force on each skateboarder is F = + 70 N + (-70 N) = + 70 N - 70 N = 0 N.

Since force, F = ma where a = acceleration and m = mass,

a = F/m.

So, since for each skater, F = 0N,

a = 0 N/m

= 0 m/s²

So, the acceleration of each skater is 0 m/s²

5 0
3 years ago
Josh has a toy car of mass 3 kg tied to a string of length 2 m. He ties the string to a pole and has the toy car drive in a circ
gulaghasi [49]

Part a)

Centripetal acceleration is defined as

a_c = \frac{v^2}{R}

now here we know that

v = 3m/s

R = 2 m

m = 3 kg

now from above formula we have

a_c = \frac{3^2}{2}

a_c = 4.5 m/s^2

Part b)

Maximum possible tension in the string is given as

T = 50 N

now by force equation we have

F = ma

50 = 3 a

a = \frac{50}{3} m/s^2

now again by above formula

\frac{v^2}{R} = \frac{50}{3}

v = \sqrt{\frac{50 \times 2}{3}}

v = 5.77 m/s

5 0
3 years ago
Read 2 more answers
Humans impact the Earth in good AND bad ways. <br><br>A) True <br><br>B) False
const2013 [10]

Answer:

True

Explanation:

yes we can see that we are helping animals but we create pollution which is very bad

7 0
3 years ago
A man runs 1200m on a straight line in 4 min . find his velocity.
luda_lava [24]

Answer:

5m/sec^2

Explanation:

Distance=1200m

Time=4 min

1=60sec

4=4 x 60

=240sec

Velocity=Distance/Time

Velocity=1200/240

Velocity=5m/sec^2

Mark me as brainliest

7 0
3 years ago
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
3 years ago
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