Question

pls solve it's urgent..i will mark your answer as brainliest​

Answer 1

Answer:

$\boxed{\sf Stress = 3.18 \times 10^8 \ Pa}$

Given:

Area of cross-section (A) = $\sf 3.14 \times 10^{-4} \ m^2}$

Force (F) = 100 kN = 100 × 10³ N = $\sf 10^5 \ N$

To Find:

Stress acting on the rod

Explanation:

Formula:

$\boxed{ \bold{\sf Stress = \frac{Force \ (F)}{Area \ of \ cross-section \ (A)}}}$

Substituting values of F & A in the equation:

$\sf \implies Stress = \frac{ {10}^{5} }{3.14 \times {10}^{ - 4} }$

$\sf \implies Stress = 0.318 \times \frac{ {10}^{5} }{ {10}^{ - 4} }$

$\sf \implies Stress = 0.318 \times {10}^{5 - ( - 4)}$

$\sf \implies Stress = 0.318 \times {10}^{5 + 4}$

$\sf \implies Stress = 0.318 \times {10}^{9}$

$\sf \implies Stress = 3.18 \times {10}^{8} \: Pa$

$\therefore$

Stress acting on the rod = $\sf 3.18 \times 10^{8} \ Pa$

Answer 2

Answer:

3.18 into 10*8

Explanation:

Area of cross-section (A) = $$\sf 3.14 \times 10^{-4} \ m^2}$$

Force (F) = 100 kN = 100 × 10³ N = $$\sf 10^5 \ N$$

To Find:

Stress acting on the rod

Explanation:

Formula:

$$\boxed{ \bold{\sf Stress = \frac{Force \ (F)}{Area \ of \ cross-section \ (A)}}}$$

Substituting values of F & A in the equation:

$$\sf \implies Stress = \frac{ {10}^{5} }{3.14 \times {10}^{ - 4} }$$

$$\sf \implies Stress = 0.318 \times \frac{ {10}^{5} }{ {10}^{ - 4} }$$

$$\sf \implies Stress = 0.318 \times {10}^{5 - ( - 4)}$$

$$\sf \implies Stress = 0.318 \times {10}^{5 + 4}$$

$$\sf \implies Stress = 0.318 \times {10}^{9}$$

$$\sf \implies Stress = 3.18 \times {10}^{8} \: Pa$$

$$\therefore$$

Stress acting on the rod = $$\sf 3.18 \times 10^{8} \ Pa$$