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Readme [11.4K]
3 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]3 years ago
4 0
U have to *modify it to increase its ground clearance*
kakasveta [241]3 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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A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
3 years ago
I'm not sure if the answer is A or B... someone help
Talja [164]
Its b i literally have had this exact question
8 0
3 years ago
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