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Black_prince [1.1K]
3 years ago
12

The graph below shows the distance traveled by the skateboarder on each of the different road conditions. Using the graph, deter

mine which of the roads was dry, wet, or muddy. Explain your answer using complete sentences.
Physics
1 answer:
igor_vitrenko [27]3 years ago
6 0

Answer:

Road A- dry

Road B- mud

Road C- wet

Explanation:

Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.

The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.

The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.

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5 0
3 years ago
Two gliders are on a frictionless, level air track. Both gliders are free to move. Initially, glider A moves to the right and gl
Yuliya22 [10]

Answer:

The change in momentum of both objects is the same but in opposite direction.

Explanation:

Hi there!

The momentum of the system is calculated as the sum of the momentums of each glider. The momentum of the system is conserved if no external force is acting on the objects (as in this case). That means that the initial momentum of the system is equal to the final momentum of the system.

The momentum of each glider is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass of the glider.

v = velocity.

The momentum of the system for glider A and B can be calculated as follows:

initial momentum = mA · vA + mB · vB

Where:

mA and vA = mass and velocity of glider A

mB and vB = mass and velocity of glider B

Initially, glider B is at rest so that vB = 0. Then, the initial momentum of the system is:

initial momentum = mA · vA

The final momentum of the system is calculated as follows:

final momentum = mA · vA´ + mB · vB´

Where vA´ and vB´ are the final velocities of glider A and B respectively.

We know that mB = 4mA and that vA´ is negative. The the final momentum will be:

final momentum = -mA · vA´ + 4mA · vB´

Since initial momentum = final momentum:

mA · vA = -mA · vA´ + 4mA · vB´

mA · vA + mA · vA´ = 4mA · vB´

<u>vA + vA´ = 4 vB´</u>

<u />

The change in momentum of glider A (ΔpA) is calculated as follows:

ΔpA = final momentum - initial momentum

ΔpA =  -mA · vA´ - mA · vA = -mA (vA + vA´) = -4mA · vB´

The change in momentum of glider B (ΔpB) is calculated as follows:

ΔpB = final momentum - initial momentum

ΔpB = 4mA · vB´ - 0 = 4mA · vB´

Then, the change in momentum of both objects is the same but in opposite direction. That´s why the momentum is conserved.

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2 years ago
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2 years ago
Read 2 more answers
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