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slamgirl [31]
3 years ago
11

A 20600 kg sailboat experiences an eastward force 17700 N due to the tide pushing its hullwhile the wind pushes the sails with a

force of 60000 N directed toward the northwest (45◦ westward of North or 45◦ northward of West). What is the magnitude of the resultant acceleration of the sailboat?
Physics
1 answer:
gavmur [86]3 years ago
3 0

Answer:

2.23 m/s²

Explanation:

mass of sail boat, m = 20600 kg

Force along east, F1 = 17700 N

Force along north west, F2 = 60000 N

write the forces in vector form

\overrightarrow{F_{1}}=17700 \widehat{i}

\overrightarrow{F_{2}}=60000\left ( - Cos45 \widehat{i}+Sin45\widehat{j}\right )

\overrightarrow{F_{2}}=-42426.41\widehat{i}+42426.41\widehat{j}

Net force acting on the boat

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=17573.6\widehat{i}+42426.41\widehat{j}

The magnitude of net force is given by

F=\sqrt{17573.6^{2}+42426.41^{2}}

F = 45922 N

Acceleration = force / mass

a = 45922 / 20600

a = 2.23 m/s²

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Answer:

For the point inside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

We have to integrate it over the ring, which is an angular integration.

E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

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A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them
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Answer:

B. There is a direct proportion between the mass and force listed in the table.

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This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

This means there is a direct proportionality relation between mass and force in the table.

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Answer:

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