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3 years ago

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A hovercraft is being driven across a lake on a very windy day. The wind exerts a force of 5000 N north on the hovercraft. The p

ropellers driving the hovercraft also exert a force.
What is the net force (to 3.s.f ) on the hovercraft when the force applied to the hovercraft by the propellers is: 10 000 N west

1 answer:

krok68 [10]3 years ago

4 0

Answer:

The net force on the hovercraft is 11200 N.

Explanation:

Given;

force exerted on the hovercraft by wind, F₁ = 5000 N north

force exerted on the hovercraft by the propeller, F₂ = 10,000 N west

The net force on the hovercraft is calculated as;

$F_{net} = \sqrt{F_1^2 + F_2^2} \\\\F_{net} = \sqrt{5000^2 + 10,000^2} \\\\F_{net} = 11180.34 \ N\\\\F_{net} = 11200 \ N \ (3.s.f)$

Therefore, the net force on the hovercraft is 11200 N.

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2) Ratio Vcenter/Vsurface: 3/2

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1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

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$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

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The potential at the surface, V(R), is that of a point charge, so

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$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

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Therefore the potential at the center of the sphere is:

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On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

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The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

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