All categories

3 years ago

East-West orbits are best for:

Physics

1 answer:

xenn [34]3 years ago

5 0

Answer: radio and television

Satellite orbits with an East-West orientation are best used for communication purposes. The satellite orbit should always be suited for its purpose. Radio and television are powered by numerous satellites rotating the Earth from east to west.

You might be interested in

I WILL GIVE BRAINLIEST

lubasha [3.4K]

Answer:

The answer is

A.Speed

and

D. Acceleration

7 0

3 years ago

Read 2 more answers

This experiment, you compare several unknown powders to known powders. What is the name given to all of the known materials used

brilliants [131]

<span>B. Controls: The known powders attributes are already studied, so the are used as control measures to compare with the findings in the experiment with the unknown powders.</span>

5 0

3 years ago

Read 2 more answers

This is an interaction in which two waves meet, matching crest to crest and trough to trough. The resultant wave will have an am

mart [117]

Answer:

jao es fac esi ylo saa lbes

Explanation:

hkhhhhkl

4 0

3 years ago

Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.

Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

$f_n = nf_1$

where

n is the number of the harmonic

$f_1$ is the fundamental frequency

Here we have:

$f_3=512 Hz$ is the frequency of the 3rd harmonic

So the fundamental frequency is

$f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz$

And so, the frequency of the 2nd harmonic is:

$f_2=2f_1=2(170.7)=341.3 Hz$

3 0

4 years ago

Fig. 2.1 shows a train

Vitek1552 [10]

Answer:

$\mathrm{(a)\:}32,000,000\:\mathrm{Ns},\\\mathrm{(b)\:}390,000\:\mathrm{N}$

Explanation:

The impulse-momentum theorem states the impulse on an object is equal to the change in momentum of that object. Momentum is given by $p=mv$ . Since mass is constant, the train's change in momentum is:

$\Delta p=m\Delta v=750,000\cdot42=31,500,000=\fbox{$32,000,000\:\mathrm{Ns}$}$ (two significant figures).

Impulse is also given as $\Delta p = F\Delta t$ , where $F$ is the average force applied and $\Delta t$ is change in time. Since $t$ is given as $80\mathrm{s}$ , we have the following equation:

$F\Delta t=\Delta p\\\\F=\frac{\Delta p}{\Delta t},\\\\F=\frac{31,500,000}{80},\\\\F=393,750=\fbox{$390,000\:\mathrm{N}$}$ (two significant figures).

7 0

3 years ago