Ask question

Ask question

All categories

Oduvanchick [21]

3 years ago

10

A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal

length, meet at an angle of 39 degrees. How long are they?

1 answer:

kogti [31]3 years ago

4 0

Answer:

The answer to your question is 636.6 ft

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

sin 19.5 = 212.5/hyp

solve for hyp

hyp = 212.5 / sin 19.5

Result

hyp = 212.5/ 0.333

hyp = 636.6 ft

You might be interested in

Explain why more stars are circumpolar for observers at higher latitudes.

PIT_PIT [208]

Explanation:

When you observe the night sky you will notice that the stars are moving. They rise from eastern horizon and set in the western horizon. It happens due to rotation of Earth. When observed closely you will notice that the all the stars seem to go around the pole star. Out of all the stars there are some stars which neither set not rise, such stars are called as Circumpolar stars. This means that they are always above the horizon. If we trace the path of such stars they will appear to make complete circle around the pole star.

Also, you will notice that the altitude of pole star (separation of pole star from the horizon in degrees) will depend on the location of observe on the Earth. This happens due to Earth being spherical. So if you are on equator the pole star will be on the horizon i.e. 0° altitude. If you are at Poles, altitude of the pole star will be 90°. Technically the altitude of pole star at any place on Earth is equal to the latitude of the place.

If the altitude of pole star varies and increases as you move towards higher latitude on Earth, the distance between horizon and pole star will also increase. This will result in more stars being circumpolar.

If you are at Poles, all the stars will be circumpolar and if you are at equator no star will be circumpolar.

8 0

3 years ago

A person is lying on a diving board 3.50 m above the surface of the water in a swimming pool. She looks at a penny that is on th

sergij07 [2.7K]

Answer:

1.995 m

Explanation:

Distance of penny as seen by the person = 5 m

Height of person from water surface = 3.50 m

Apparent depth of penny = 5 - 3.50 = 1.5 m

refractive index of water, n = 1.33

real depth / apparent depth = n

real depth = 1.33 x 1.5 = 1.995 m

Thus, the actual depth of water at that point is 1.995 m.

4 0

3 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago

Which of the following is an example of a force<br> inertia<br> push <br> pull

Whitepunk [10]

A force can be considered a push or pull

hope this helps :)

8 0

3 years ago

List the steps you would use to reach one of your personal growth goals.

Zinaida [17]

Im not sure, but here are a few of mine : Learn to be silent, and listen, take massive action and be proactive, listen, focus, and lastly, persist. Hope this helped! (:

6 0

3 years ago