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elena55 [62]
3 years ago
7

Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow

n along the plane. After traveling some distance down the inclined plane it comes to rest due to frictional forces. In the simplest case, there are three forces acting upon the motion of the block—the initial push, gravity, and friction. Which of these statements about the work done by these three forces is true?
-The work done by gravity is equal to the sum of the work done by friction and the initial push.

-All three forces contribute an equal amount of work.

-The work done by friction is equal to the sum of the work done by gravity and the initial push.

-The work done by the initial push is equal to the sum of the work done by friction and gravity.
Physics
1 answer:
katen-ka-za [31]3 years ago
6 0

Answer:

the correct affirmation is the 3

Explanation:

Let's analyze the problem with Newton's second law before looking at the claims.

X axis parallel to the plane, positive down

      F -fr + Wₓ = ma

Y Axis perpendicular to the plane

      N -Wy = 0

With trigonometry

     Wₓ = W sin θ

     Wy = w cos θ

Let's multiply by the displacement along the plane, to relate to the work, which has as expression W = F d

     F d -fr d + Wx d = ma d

Push                  W₁ = Fd

frictional force  W₂ = -fr d

gravity               W₃ = Wx d

    W₁ + W₃ -W₂ = m a d

Analysis affirmations:

R1) false. The work of gravity is the subtraction    

R2) false. Each force contributes according to its magnitude

R3) true. In the equation we see that, if the acceleration is zero, W2 = W1 + W3

R4) False. It equals the difference

the correct affirmation is the 3

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ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

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She uses a voltmeter, that measures in volts, and an ammeter that measures in amps. Both were correctly placed in her circuit. I
Nesterboy [21]

Answer:

A) the ammeter is x  

B)

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  • voltage across the right one  = 0.3 V

C) 1.05 V

Explanation:

From the diagram attached below;

A) Assuming the homes were wired in series, and one of the homes face short circuit then all the houses would face power cut but it doesn't happen. So they must be connected in parallel.

Therefore; The ammeter is  connected in series, Hence, the ammeter is x  and the voltmeter must be z.

B)

Given that:

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= 5×0.15

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C)

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