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1 year ago

A child has a bucket full of toys: foam alphabet letters, action figures, and toy cars.

2 answers:

7 0

the answer is density because the foam letters are denser meaning they will float to the surface quicker than the other toys

VLD [36.1K]1 year ago

7 0

Answer: B. Density

Explanation:

Density is the comparison between the amount of matter contained by an object and its volume.

Mathematically, density is defined as; d = m/v

Where d is density

m is mass

and v is volume.

Dumping the toys in the bathwater is taking advantage of density because foam is generally less dense than water and anything less dense than water will float in water.

So, there will be a separation based on density because the foam alphabet letters will float while the other toys will sink due to the fact that they are denser and heavier than water and their molecules are compact; very close together.

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If a magnet is broken into two pieces, what happens to the magnetic poles? One piece will have a north pole, while the other pie

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Each piece will still have a North and South Pole

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A source charge generates an electric field of 4286 N/C at a distance of 2.5 m. What is the magnitude of the source charge?

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The Answer is 3.0uc. I took the quiz.

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1 year ago

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At a local swimming pool, the diving board is elevated h = 5.5 m above the pool's surface and overhangs the pool edge by L = 2 m

Margaret [11]

Answer:

Part a)

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = 1.06 s$

Part c)

$L = 4.86 m$

Explanation:

Part a)

The height of the diving board is given as

$h = 5.5 m$

now the speed of the diver is given as

$v_0 = 2.7 m/s$

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board

So we will have

$y = v_y t + \frac{1}{2}at^2$

$h = 0 + \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}}$

Part b)

$t = \sqrt{\frac{2h}{g}}$

plug in the values in the above equation

$t = \sqrt{\frac{2(5.5 m)}{9.81}$

$t = 1.06 s$

Part c)

Horizontal distance moved by the diver is given as

$d = v_0 t$

$d = 2.7 \times 1.06$

$d = 2.86 m$

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1 year 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:

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

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

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

Explanation:

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

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

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

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

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

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)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

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

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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