Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant

Physics

1 answer:

Cerrena [4.2K]2 years ago

4 0

Answer:

the force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

$F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81 + \frac{70 \times 120^2}{100}\\\\= 10766.7 N$

Hence, the force exerted by the seat on the pilot is 10766.7 N

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A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla

MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, $u=3\times 10^5\ m/s$

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

$qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s$