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1 year ago

11

A "sound" is different than a "sound wave"

1 answer:

e-lub [12.9K]1 year ago

5 0

Answer:

sound is how an animal or human perceives a sound wave

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What is the difference between Drift current & Diffusion current ?

dsp73

The main difference between the drift current and diffusion current is the corresponding reaction of the semiconductor materials in the electric fields. In electric fields, when semiconductor material's movement go along with the direction of the electric field it is called drift current, but when it tends to go from high concentrated area to lower concentrated area then it is diffusion current.

8 0

3 years ago

A point charge Q is located a short distance from a point charge 3Q, and no other charges are present. If the electrical force o

weeeeeb [17]

Answer:

d) F

Explanation:

According to columb's law:

"The magnitude of electrostatic force between two charges is directly proportional to the product of magnitude of two charges and inversly proportional to separation between them."

If q₁ and q₂ are magnitude of two charges, d is distance between them and k is dielectric constant, then force F is given by

$F=\frac{kq_{1}q_{2}}{d^2}$

According to this force exerted on point charge Q is same as that of 3Q, so force point 3Q charge experience is also F

7 0

2 years ago

A teacher performing demonstration finds that a piece of cork displaces 23.5 ml of water. The piece of cork has a mass 5.7 g. Wh

Sedbober [7]

Answer: 0.24g/ml

Explanation:

Given that:

Volume of water displaced = 23.5 ml

Mass of cork = 5.7 g

Density of the cork = ?

Recall that density is obtained by dividing the mass of a substance by the volume of water displaced.

i.e Density = Mass/volume

Density = 5.7g /23.5ml

Density = 0.24g/ml

Thus, the density of the piece of cork is 0.24g/ml

8 0

3 years ago

A ski jumper starts at the top of a 400m high hill before skiing down to the bottom. If the ski jumper has a mass of 75kg how mu

alekssr [168]

Answer:B. 30,000 J

Explanation:

6 0

2 years ago

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

2 years ago