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Found in: Page 396

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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

The Pressure exerted on shot-put is obtained as: $1.6×{10}^{5}Pa$.

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

See the step by step solution

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 $F=800N$.

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:

$P=\frac{F}{A}\phantom{\rule{0ex}{0ex}}=\frac{800}{0.005}\phantom{\rule{0ex}{0ex}}=160000 \text{Pa}\phantom{\rule{0ex}{0ex}}$

The Pressure exerted on shot-put is: $1.6×{10}^{5}Pa$.

Relation of the pressure in $1×{10}^{6}Pa$.

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

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

$Ratio=\frac{1.6×{10}^{5}}{1×{10}^{6}}\phantom{\rule{0ex}{0ex}}=0.16\phantom{\rule{0ex}{0ex}}$

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

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