PHYSICS OF EVERYDAY PHENO... 7/14 >C<
PHYSICS OF EVERYDAY PHENO... 7/14 >C<
8th Edition
ISBN: 9781308172200
Author: Griffith
Publisher: MCG/CREATE
Question
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Chapter 9, Problem 1SP

(a)

To determine

The area of input and load piston.

(a)

Expert Solution
Check Mark

Answer to Problem 1SP

The area of input piston is 3.14cm2 and that of load piston is 490.87cm2.

Explanation of Solution

Given info: The input piston has diameter of 2 cm and the output piston has diameter 25 cm. The mass of the car is 1400 kg.

The radius of the input piston is 1 cm and the radius of the output piston is 12.5 cm.

Write the equation to find the area of input piston.

Ai=πr2

Here,

Ai is the area of input piston

r is the radius of the input piston

Substitute 1 cm for r in equation to find the area of the input piston.

Ai=π(1 cm)2=3.14cm2

Write the equation to find the area of load piston.

Al=πr2

Here,

Al is the area of load piston

r is the radius

Substitute 12.5 cm for r in equation to the area of the load piston.

Al=π(12.5 cm)2=490.87cm2

Conclusion:

Therefore, the area of input piston is 3.14cm2 and that of load piston is 490.87cm2.

(b)

To determine

The ratio of area of load piston to area of input piston.

(b)

Expert Solution
Check Mark

Answer to Problem 1SP

The ratio of area of load piston to area of input piston is 156.25:1.

Explanation of Solution

Write the equation to find the ratio of area of load piston and input piston.

ratio=AlAi

Substitute 490.87cm2 for Al and 3.14cm2 for Ai to get the ratio.

ratio=490.87cm23.14cm2=156.25

Conclusion:

Therefore, the ratio of area of load piston to area of input piston is 156.25:1.

(c)

To determine

The weight of car in newtons.

(c)

Expert Solution
Check Mark

Answer to Problem 1SP

The weight of car in newtons is 1.372×104N.

Explanation of Solution

Write the equation to find the weight.

W=mg

Here,

W is the weight

m is the mass

g is the acceleration due to gravity

Substitute 1400 kg for m and 9.8m/s2 for g in equation to find the weight.W=(1400 kg)(9.8m/s2)=1.372×104N

Conclusion:

Therefore, the weight of car in newtons is 1.372×104N.

(d)

To determine

The force applied to piston to support the car.

(d)

Expert Solution
Check Mark

Answer to Problem 1SP

The force applied to piston to support the car is 87.8N.

Explanation of Solution

Write the expression to find the pressure needed on the input piston to support the car of weight 1.372×104N.

P=FAi

Here,

F is the force corresponding to the weight of the car

Substitute 1.372×104N for F and 490.87×104m2 for Al in the above equation to find the pressure needed on the input piston to support the car of weight 1.372×104N.

P=1.372×104N490.87×104m2=2.79×103 Pa

Write the equation to find the force applied to piston to support the car.

F=PA

Here,

P is the pressure needed on the input piston to support the car of weight 1.372×104N

F is the force

A is the area

Substitute 2.79×103 Pa for P and 3.14×104m2 for A in equation to find the force applied to piston to support the car.

F=(2.79×103 Pa)(3.14×104m2)=87.76N87.8N

Conclusion:

Therefore, the force applied to piston to support the car is 87.8N.

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Chapter 9 Solutions

PHYSICS OF EVERYDAY PHENO... 7/14 >C<

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