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RoseWind [281]
3 years ago
10

how much force (in N) is exerted on one side of an 16.2 cm by 22.9 cm sheet of paper by the atmosphere

Physics
1 answer:
luda_lava [24]3 years ago
6 0
<span>The 16.2cm by 22.9cm paper has an area = 370.98 square cm = 0.037098 square meters. And the atmospheric pressure at sea level is 101,325 Newtons per square meter. So if multiply the quantities: 101,325 N/m^2 and 0.037098 m^2 we get an answer of 3758.95 N. * If we respect the fact that 22.9cm and 16.2 cm only have three significant digits then our answer would be 3760N.</span>
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Answer:

The force is the same

Explanation:

The force per meter exerted between two wires carrying a current is given by the formula

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 is the vacuum permeability

I_1 is the current in the 1st wire

I_2 is the current in the 2nd wire

r is the separation between the wires

In this problem

I_1=2.79 A\\I_2=4.36 A\\r = 9.15 cm = 0.0915 m

Substituting, we find the force per unit length on the two wires:

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(2.79)(4.36)}{2\pi (0.0915)}=2.66\cdot 10^{-5}N

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.

The same conclusion comes out  from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).

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Explanation:

Given that,

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\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

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For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

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