Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is ${50.0 cm}^{2}$ and he exerts a force of role="math" localid="1668682357057" $800 N$ on it. Express the pressure in and compare it with the $1.00 \times 10^{6} P a$ pressures sometimes encountered in the skeletal system.

The Pressure exerted on shot-put is obtained as: $1.6 \times 10^{5} P a$ .

The ratio of both the values is: $0.16$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Conceptual Introduction

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

Step 2: Given Data

The pressure is the rate of force applied to the surface per unit area.

The Force which is given by the hand of shot-putter to the shot-put is $F = 800 N$ .

The area of the contact in meter square is $A = 0.005 m^{2}$ .

Step 3: Pressure exerted on the shot put

Since they are equivalent, we can write the above equation as:

$P = \frac{F}{A} = \frac{800}{0.005} = 160000 Pa$

The Pressure exerted on shot-put is: $1.6 \times 10^{5} P a$ .

Relation of the pressure in $1 \times 10^{6} P a$ .

As we know $1 N / m^{2} = 1 P a$ .

Hence, we need to do ratio to compare this both quantity

$R a t i o = \frac{1.6 \times 10^{5}}{1 \times 10^{6}} = 0.16$

Therefore, the ratio of the two value is $0.16$ .

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Conceptual Introduction

Step 2: Given Data

Step 3: Pressure exerted on the shot put

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Conceptual Introduction

Step 2: Given Data

Step 3: Pressure exerted on the shot put

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