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Maurinko [17]
3 years ago
10

Help me out!! please!

Physics
1 answer:
AVprozaik [17]3 years ago
7 0
42- C
43- A
44- C
45- A
46- B
47- D
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The angle of incidence of a ray of light striking an equilateral triangular prisms ABC of refracting angle 60o is 40o. Calculate
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Answer:

1: the refracted angle in the first face is equal to the incident angle that is 60degrees

2. Emergence Angle is 42degrees

Explanation:

Pls see attached file

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What conditions are required for a solar eclipse?
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The phase of the Moon must be new, and the nodes of the Moon's orbit must be nearly aligned with Earth and the Sun.
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Moving company uses a machine to raise a 900 Newton refrigerator to the second floor of a building machine consists of a single
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Mechanical advantage of a machine is the ratio of the output force over the input force or M=Fo/Fi. Since M=1, Fi=Fo, or the input force is equal to the output force. This means that to raise the refrigerator that weighs 900 N, we need the same input force of 900 N, or Fo=Fi=900 N. 
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3 years ago
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4.5 cm3 of water is boiled at atmospheric pressure to become 4048.3 cm3 of steam, also at atmospheric pressure. Calculate the wo
Mrrafil [7]

1. 408.4 J

The work done by a gas is given by:

W=p\Delta V

where

p is the gas pressure

\Delta V is the change in volume of the gas

In this problem,

p=1.01\cdot 10^5 Pa (atmospheric pressure)

\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3 is the change in volume

So, the work done is

W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J

2. 10170 J

The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:

Q = m \lambda_v

where

m is the mass of the water

\lambda_v = 2.26\cdot 10^6 J/kg is the specific latent heat of vaporization

The initial volume of water is

V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3

and the water density is

\rho = 1000 kg/m^3

So the water mass is

m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg

So, the amount of heat added to the water is

Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J

8 0
4 years ago
PLEASE HELP WILL MAKE BRAINIEST (GIVING AWAY 20 POINTS) Where on the track is the most potential energy? Position A Position B P
Vladimir [108]

Answer:

Though there is no chart on my screen, I can give you the correct information to solve the question without knowing what it looks like. Assuming that this track has slopes and or ramps of any kind, the place where it would have the most potential energy would be at the highest point of this ramp, or and the highest point on the track. This would also mean that it would fall the farthest. The highest kinetic energy would be at the lowest part of the ramp, after its used the slope to its full ability. But this would also be before it hits flat ground, or starts going up another ramp. This would be the point where it would be going its fastest.

Explanation:

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3 years ago
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