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Readme [11.4K]
4 years ago
10

You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when

you are finished
Physics
2 answers:
jok3333 [9.3K]4 years ago
4 0
U have to *modify it to increase its ground clearance*
kakasveta [241]4 years ago
3 0

You have a pick-up truck that weighed 4,000 pounds when it was new. You are modifying it to increase its ground clearance. You are finished when you increase the vehicles bumper height and reduce anything that is not needed to lower the weighted.

<h3>Further explanation </h3>

The gross weight average of a small pickup truck is between 5,000 and 7,000 pounds. Besides the heavy duty pickups weigh as much as 7,500 to 12,000 pounds

t is not advisable to increase the ground clearance of the vehicle becuaseit affects the vehicle stability and also driving dynamics. It is advised only if you are in desperate need of higher ground clearance.

Before you start to modify the truck, make sure about why you are making the change, what you are hoping to pick up and what impact it will have on the whole vehicle.

One way to increase the ground clearance of truck is to use spacers under the coil spring. This will affect the overall handling and driving as the angle of the lower arms and links will change, therefore the ride will become more stiff.

<h3>Learn more</h3>
  1. Learn more about pick-up truck brainly.com/question/1148453
  2. Learn more about ground clearance brainly.com/question/3539288
  3. Learn more about showroom vehicle brainly.com/question/4521658

<h3>Answer details</h3>

Grade:  9

Subject:  physics

Chapter:  weight

Keywords:  pick-up truck, ground clearance, truck, tires, showroom vehicle

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
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Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

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F = k \frac{q_{e}q}{r}

where

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since it is measured at the midpoint,

r = \frac{0.5}{2}

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F = F_{1}+ F_{2}

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the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

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