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OverLord2011 [107]
1 year ago
11

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

me have more details available when you click on them. Drag and drop each label to the corresponding area it identifies in the image.
ITEM BANK: Move to Bottom
(I'm in 8th grade and I seriously just need a answer to this)

Physics
1 answer:
Vikki [24]1 year ago
7 0
1 isla thanks for your help and I hope you are feeling 44356
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The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
2 years ago
Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288
Tomtit [17]
<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football. 
</span>
7 0
2 years ago
Read 2 more answers
(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres
timurjin [86]

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

5 0
1 year ago
B. Complete the table to show the effect of each change on each electric quantity. (1 point)
notka56 [123]

Answer:

Effect on electric force

Multiply one charge by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is directly proportional to the charge.

Hence if one charge is doubled, then the electric force is doubled.

Multiply distance by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is inversely proportional to the square of the distance of separation.

If the distance is doubled, F is decreased by 22. This means that the force is multiplied by 1/4.

Effect on electric potential energy

Multiply one charge by 2.

The electric potential energy is given by U=kq1q2/r

From the equation, the electric potential energy is directly proportional to the charge q.

If one charge is doubled, the electric potential energy is doubled.

Multiply distance by 2

The electric force is given by U=kq1q2/r

From the equation, the electric potential energy is inversely proportional to the distance of separation r.

If the distance is doubled, U is divided by 2. This means that the electric potential energy is multiplied by 1/2.

Effect on potential difference

Potential difference is defined as the change in electric potential energy.

Increase in the charge causes an increase in the potential difference and an increase in the distance of separation decreases the potential difference.

4 0
2 years ago
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