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Maksim231197 [3]

2 years ago

12

When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying

its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period

1 answer:

hammer [34]2 years ago

5 0

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ax$ (1)

Where:

x: is the displacement

$v_{f}$ : is the final speed = 0 (reduces its velocity back to zero)

$v_{0}$ : is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

$x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m$

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

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Explanation:

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We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

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c)

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