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julsineya [31]
3 years ago
8

In the lab setup above, block 1 is being pulled across a smooth surface by a light string connected across a pulley to a falling

block 2. The pulley has negligible mass and friction. Students measure the mass of each block, the distance from A to B , and the speeds of the block on the surface at points A and B . Which of the following hypotheses can be tested with this setupAs block 1 moves from point A to point B, the work done by the tension on block 1 is equal to the kinetic energy of block 1 at point B.As block 1 moves from point A to point B, the work done by the tension on block 1 is equal to the kinetic energy of the two-block system when block 1 is at point B.As block 1 moves from point A to point B , the work done by gravity on block 2 is equal to the kinetic energy of block 1 at point B.As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.As block 1 moves from point A to point B, the work done by the tension on block 2 is equal to the kinetic energy of block 2 when block I has reached point B.
Physics
1 answer:
Nuetrik [128]3 years ago
8 0

Answer:

As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.

Explanation:

As block 2 goes down , work is done by gravity on block 2 . This is converted

into kinetic energy of block 1 and block 2 . Work done by gravity is mgh which can be measured easily . kinetic energy of both the blocks can also be measured.  

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6.3445×10⁻¹⁶ m

Explanation:

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E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s

Deflection by Earth's Gravity

\Delta =\frac {gt^2}{2}

Now, Time = Distance/Velocity

\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m

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3 years ago
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5 0
3 years ago
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Change of motion are caused by
AlexFokin [52]
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4 years ago
A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .
Rzqust [24]

Answer:

a)  ΔV = 2,118 10⁻⁸ m³   b)  ΔR= 0.0143 cm

Explanation:

a) For this part we use the concept of density

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As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats

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We use the relation of the bulk module

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    ΔV = V P / B

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b) indicates that we approximate the diamond to a sphere

    V = 4/3 π R³

For this part let's look for the initial radius

    R₀ = ∛ ¾ V /π

    R₀ = ∛ (¾ 0.159 /π)

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Now we look for the final volume and with this the final radius

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The radius increment is

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4 0
3 years ago
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<u>Principal</u><u> </u><u>focus</u><u> </u><u>of</u><u> </u><u>concav</u><u>e</u><u> </u><u>lens</u><u> </u><u>-</u><u> </u>

★ The point at which rays parallel to principal axis coming from infinity appear to converge after being refracted from concave lens is called the principal focus of concave lens.

<em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em>

• <u>Additional</u><u> information</u><u> </u><u>-</u><u> </u>

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8 0
3 years ago
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