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Lana71 [14]
3 years ago
7

Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all

four of his original coil springs with new ones offered as an option by the car manufacturer. Note that changing the coil springs on a car can unsafely affect its handling. Consult your national or state laws before altering the suspension on any car. The new springs will be identical to the original springs, except the force constant will be 5855.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height?
Physics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

0.05 cm

Explanation:

The compression of the original spring = 12 - 8.55 cm = 3.45 cm = 0.0345 m

By Hooke's law, F = ke

Where F is the applied force, k is the spring constant and e is the extension or compression. In the question, F is the weight of the car.

k = F/e = 1355 × 9.8 / 0.0345 = 384898.55 N/m

This is the spring constant of the original spring. The question mentions that the force constant of the new spring is 5855.00 N/m smaller. Hence, the force constant of the new spring is 384898.55 - 5855 = 379043.55 N/m

With the new spring installed, the compression will be

e = F/k = 1355 × 9.8 / 379043.55 = 0.035 m = 3.5 cm

The difference in the compressions of both springs = 3.5 cm - 3.45 cm = 0.05 cm

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What is the acceleration of a 1000 kg car subject to a 550 N net force
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Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 ti
castortr0y [4]

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(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

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Solving (a): F = 2C

Substitute 2C for F in the above equation

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Collect like terms

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Substitute ¼C for F in the above formula

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Convert fraction to decimal

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