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pav-90 [236]
3 years ago
13

A housefly walking across a clean surface can accumulate a significant positive or negative charge. In one experiment, the large

st positive charge observed was +70 pC. A typical housefly has a mass of 12mg. A.) Explain how the housefly could accumulate an electric charge by walking across a surface? B.) What magnitude and direction electric field would be necessary to levitate a housefly with a maximum charge? Could such a field exist in air?
Physics
1 answer:
nalin [4]3 years ago
5 0

Answer:

1681714.28571 N/C

Yes it could exist

Explanation:

m = Mass of housefly = 12 m

q = Charge = 70 pC

g = Acceleration due to gravity = 9.81 m/s²

E = Electric field

When an object accumulates charge it means that it is gaining electrons making it negatively charged. This is the concept of static electricity.

Here, the electric force and the graviational force will balance each other

F_e=W\\\Rightarrow qE=mg\\\Rightarrow E=\frac{mg}{q}\\\Rightarrow E=\frac{12\times 10^{-6}\times 9.81}{70\times 10^{-12}}\\\Rightarrow E=1681714.28571\ N/C

1681714.28571 N/C of electric field would be required to levitate

The direction of the electric field would be upwards vertically.

In air the critical value of electric field is 3\times 10^6\ N/C which is more than the critical value of electric field in air. So, the electric field can exist.

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SCORPION-xisa [38]
Answer: 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).

Explanation:

1) Since the mass of <span>155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).
</span>

<span>2) You do that using the trignometric ratios sine and cosine.
</span>

<span>Weight is mass × g.
</span>
<span>Weight of the object = 155g × g
</span>
<span>Angle, α = 210°
</span>

<span>Horizontal component (0°)
</span>
<span>cosα = horizontal / hypotenuse ⇒ horizontal = hypotenuse × cosα
</span>
⇒ horizontal = 155g × g × cos(210°) = - 134.23g  × g

Vertical component
sinα = vertical / hypotenuse ⇒ vertical = hypotenuse × sinα
⇒ vertical = 155g × g × sin(210°) = -77.5g × g

3) Conclusion:

Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).




8 0
3 years ago
Kinematics in 1D [4B.R]:Question 3
Stels [109]

Answer:

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6 0
4 years ago
What is the difference between a unit and a variable?
scoundrel [369]

Answer:

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7 0
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How should you approach a dock when the wind or current is pushing you away from the dock?
erastova [34]

Answer:

If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.

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An object experiences an acceleration of 6.8m/s2.As a result,it accelerates from rest to 24m/s.How much distance did it travel d
Hitman42 [59]

The distance covered by the object is 42.4 m

Explanation:

The motion of the object is a uniformly accelerated motion (at constant acceleration), therefore we can use the following suvat equation:

v^2 -u^2 = 2as

where

v is the final velocity

u is the initial velocity

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s is the distance covered

For the object in this problem, we have:

u = 0 (it starts from rest)

v = 24 m/s (final velocity)

a=6.8 m/s^2

Solving for s, we find the distance travelled by the object:

s=\frac{v^2-u^2}{2a}=\frac{24^2-0}{2(6.8)}=42.4 m

Learn more about accelerated motion:

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