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Arada [10]
2 years ago
13

2 Grade 11 Physics Questions. Will Mark BRAINLIEST!!!

Physics
2 answers:
Stels [109]2 years ago
7 0

Answer:

For the first one it would be screw and for the second one it is a ramp

Explanation:

Hope this helps

Viktor [21]2 years ago
3 0
Thanks for the help
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Why does the current is reduced as electrons move through a conductor
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1. Los Angeles lies on Pacific plate, San Francisco lies on North American plate. When will the two cities meet if the distance
spin [16.1K]

Los Angeles lies on the Pacific plate, San Francisco lies on the North American plate, and the meeting point of the two cities is mathematically given as

T = 120 x 105 years

<h3>What is the meeting point of the two plates?</h3>

Generally, the equation for Distance is mathematically given as

D = Rate x Time

Therefore

T = D/R

T = (600 x 105) / 5

T = 120 x 105 years

In conclusion, the meeting point of the two plates will be

T = 120 x 105 years

Read more about Arithmetic

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1 year ago
Dr. Matthews has submitted a proposal to the institutional review board (IRB) of a university. At this university, she intends t
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6 0
3 years ago
A copper wire has a diameter of 2.097 mm. what magnitude current flows when the drift velocity is 1.54 mm/s? take the density of
frozen [14]

By using drift velocity of the electron, the current flow is 7.20 ampere.

We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as

v = I / (n . A . q)

where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.

From the question above, we know that

d = 2.097 mm

r = (0.002097 / 2) m

v = 1.54 mm/s = 0.00154 m/s

ρ = 8.92 x 10³ kg/m³

q = e = 1.6 x 10¯¹⁹C

Find the atom density

n = Na x ρ / Mr

where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).

n = 6.022 x 10²³ x 8.92 x 10³ / 0.635

n = 8.46 x 10²⁷ /m³

Find the current flows

v = I / (n . A . q)

0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)

0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)

I = 7.20 ampere

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1 year ago
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