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2 years ago

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In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

such that the centripetal acceleration is large enough to keep the string that. Determine the minimum speed for the ball to achieve a complete circle in terms of L, d, and g. (Hint: use Newton’s 2nd Law and consider what the string tension must be for the ball to just barely get over the top).

1 answer:

Vesna [10]2 years ago

6 0

Answer:

Explanation:

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

r=distance between point of rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

$v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

$v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}$

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A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

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Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

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$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

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2 years ago

A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

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Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

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Explanation:

Length of sample, $l_{s} = 2 cm$

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E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

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$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

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Answer:

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2 years ago

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