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2 years ago

Proper use of the friction zone makes it easier to:.

1 answer:

4 0

Friction is the opposition offered to the movement of a surface on another and depends on the nature of the surfaces in contact.

<h3>What is friction?</h3>

The term friction refers to the opposition that is offered by surfaces in contact when they move over each other. The question is incomplete but I will try to help you the much I can.

Now recall that fricton only depends on the nature of the surfaces in contact. The friction is te reason when heat is provides in the moving parts of machines.

Learn more about friction: brainly.com/question/15117038

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A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m

Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0

3 years ago

Read 2 more answers

Which of the following is closer to the glomerulus?

VLD [36.1K]

A - proximal convoluted tubule

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3 years ago

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A BODY STARTING FROM REST MOVES WITH CONSTANT ACCELERATON. what is the ratio of distance covered by the body during the fifth se

Art [367]

Answer:

9/25

Explanation:

Distance covered in the first 5 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₅ = (0) (5) + ½ a (5)²

Δx₀₋₅ = 25a/2

Distance covered in the first 4 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₄ = (0) (4) + ½ a (4)²

Δx₀₋₄ = 8a

So the distance covered during the 5th second is:

Δx₅ = 25a/2 − 8a

Δx₅ = 9a/2

So the ratio of the distance covered during the 5th second to the distance covered in the first 5 seconds is:

Δx₅ / Δx₀₋₅

(9a/2) / (25a/2)

9/25

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3 years ago

A 0.40-kg particle moves under the influence of a single conservative force. At point A, where the particle has a speed of 10 m/

yan [13]

Answer:

The value of the potential energy of the particle at point B is 85 joules.

Explanation:

According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:

$K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}$