Question

Physics is as easy as falling off a log :')

Solve the question given in the attachment.

Answer:- Option A.

Kindly explain it _/\_
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Answer 1

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We know :

$\bullet\ \sf{| \overrightarrow{\Delta P}|=2 mvcos\ \theta(\Delta t)}$

$\sf{F=\dfrac{\overrightarrow{\Delta P}}{\Delta t} }\\\\\sf{\longmapsto F=2mv\ cos\theta }$

We also know,

$\bullet\ \sf{P=\dfrac{F}{A} }$

$\sf{\longmapsto P=\dfrac{2mv\ cos \theta }{A} }$

$\sf{\longmapsto Net\ Pressure=NP=\dfrac{2mv\ cos \theta\ N}{A} }$

We have :-

• Mass, m = 3.32 × 10⁻²⁷ kg
• N = 10²³
• cos θ = cos 45° = 1/√2
• Velocity, v = 10³ m/s
• Area, A =  2 cm² = 2 × 10⁻⁴ m²

Substituting all the values in the equation above :-

$\sf{\longmapsto P=\dfrac{x2 \times 3.32 \times 10^{-27}\times 10^2^3\times 10^3 \times \frac{1}{\sqrt{2} } }{2 \times 10^-^4} }$

$\displaystyle \sf{\longmapsto P=\frac{2 \times \frac{1}{\sqrt{2} }\times 3.32 \times 10^{-27}\times 10^{23}\times 10^3 \times 10^4 }{2} }\\\\ \displaystyle \sf{\longmapsto P=\frac{\sqrt{2} \times 3.32 \times 10^{-27+23+4}\times 10^3}{2} }\\\\\displaystyle \sf{\longmapsto P=\sqrt{2} \times 1.66 \times 10^0 \times 10^3}\\\\\displaystyle \sf{\longmapsto P=1.414 \times 1.66 \times 1 \times 10^3}\\\\\displaystyle \sf{\longmapsto P=2.347 \times 10^3}$

$\boxed{\displaystyle \bf{\longmapsto P=2.35 \times 10^3\ N/m^2}}$

Therefore, the answer is (1) 2.35 × 10³ N/m².

Answer 2

Answer:

2.35 × 10³

Explanation:

Please consider the attached picture.