A force of $600 N$ acts on the bullet of the area of cross section $0.6 {c m}^{2}$ . The pressure exerted is $10^{x}$ . What is X?

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Solution

$A r e a = 6.6 {c m}^{2} = 0.6 \times 10^{- 4} m^{2}$
Force = 600N.
So, Pressure = $\frac{F o r c e}{A r e a} = \frac{600}{0.6 \times 10^{- 4}} P a$
Pressure = $10^{7} P a$

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A force of 600N acts on the bullet of the area of cross section 0.6cm2. The pressure exerted is 10x. What is X?

Area=6.6cm2=0.6×10−4m2Force = 600N.So, Pressure = ForceArea=6000.6×10−4Pa Pressure = 107Pa

A force of $600 N$ acts on the bullet of the area of cross section $0.6 {c m}^{2}$ . The pressure exerted is $10^{x}$ . What is X?

$A r e a = 6.6 {c m}^{2} = 0.6 \times 10^{- 4} m^{2}$
Force = 600N.
So, Pressure = $\frac{F o r c e}{A r e a} = \frac{600}{0.6 \times 10^{- 4}} P a$
Pressure = $10^{7} P a$