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1 year ago

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A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at

0 degrees celsius. The aluminum ring ( a = 2.4 x 10^-5/degrees celsius) has an inside diamter of 3.994 cm at 0 degrees celsius. At what temperature will teh sphere fall through teh ring??Mulitple Choice Questiona, 461.5 degrees celsiusb. 208.0 degrees celsiusc. 115.7 degrees celsiusd. 57.7 degrees celsius

1 answer:

mote1985 [20]1 year ago

8 0

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

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Explanation:

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In other words, the algebraic sum of all the potential differences through a loop must be equal to zero.

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2 years ago

A charge of 4 nc is placed uniformly on a square sheet of nonconducting material of side 17 cm in the yz plane. (a) what is the

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The charge density of the sheet is 1.384×10⁻⁷C/m².

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Substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

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2 years ago

It takes a slug 20 minutes to travel from the grass to the trash can , a trip of 15 meters. How far could the slug travel in 60

inn [45]

Answer:

45 meters

Explanation:

20 min = 15 meters

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2 years ago

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Answer:

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the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

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ΔE = no of fixture × number of lamp × power of each lamp × Δt

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8 0

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

Airida [17]

Answer:

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Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

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Separation between the two probes, L = 1.8 cm

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Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

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and the Electric field strength, $E = \frac{V}{l_{s} }$

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E = 2.5 V/cm

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

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.

4 0

2 years ago