Pls help it's due today

A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of h

mrs_skeptik [129]

Answer:

1199 miles

Explanation:

1 hour 30 minutes = 1 + 30/60 = 1.5 hours

2 hours 15 minutes = 2 + 15/60 = 2.25 hours

The distance she flew in the 1st segment is:

1.5*345 = 517.5 miles

The distance she flew in the 2nd segment is:

2.25 * 345 = 776.25 miles

Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown

776.25 * cos(45) = 549 miles in-line with the 1st segment and

776.25* sin(45) = 549 miles perpendicular to the 1st segment:

So the distance from the end to her starting position is

$\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles$

4 0

2 years ago

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

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.

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.

We will only change the boundaries of the last integration.

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

6 0

2 years ago

A bicycle wheel has spokes for support. Each spoke extends from the center of the wheel to the rim. Which method can be used to

hoa [83]

Since the formula for the circumference of a circle is pi x d or pi x 2r. The radius is a straight line from the center of the circle to the outer part of it, which is exactly what the spokes are to the wheel. So we take the length of the spoke, which is the radius, and use it in our equation. pi times 2 times the length of the spoke will give us our answer.

3 0

3 years ago

Read 2 more answers

Make a quantitative graph for the motion of the two cars

Andru [333]

It would be helpful if you gave me a bit more information on what the cars speed is

6 0

2 years ago

How much heat is absorbed by a 450g gold block as energy from the sun causes its temperature to change from 27°C to 32°C? (Speci

kotegsom [21]

Answer:

Q = 1057.5 [cal]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

$Q=m*C_{p}*(T_{final}-T_{initial})$

where:

Q = heat energy [cal]

Cp = specific heat = 0.47 [cal/g*°C]

T_final = final temperature = 32 [°C]

T_initial = initial temperature = 27 [°C]

m = mass of the substance = 450 [g]

Now replacing:

$Q=450*0.47*(32-27)\\Q=1057.5[cal]$

5 0

2 years ago

Pls help it's due today​

Pls help it's due today