Why would scientists use a hand lens to observe minerals?

An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit

zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

$F=qE$

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

$q=3.0\cdot 10^{-6}C$

$E=-1.5\cdot 10^4 N/C$ (negative because inward)

So the strength of the electrostatic force is

$F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N$

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0

4 years ago

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What is jupiters orbital period

padilas [110]

11.86 years. Usually memorized as "12 years".

8 0

3 years ago

The Nazca Seafloor Plate Pushes Into The South American Continental Plate. What Is The Most Likely Result?

Anika [276]

The Nazca plate will move under the south american plate. i know its late but it will help others!

3 0

3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

Explain why radio and television stations use different frequencies to broadcast programs.

Sedaia [141]

I'm not sure exactly what you're asking.

It could be

==> Why do radio stations use different frequencies from TV stations ?

or it could be

==> Why don't all radio stations, or all TV stations, use the same frequency ?

Radio and TV can't coexist among each other in the same "band"

of frequencies, because they use different amounts of "space" on

the dial. One analog TV channel uses enough dial space for about

600 AM radio stations, or 30 FM radio stations ! That's one click on

the TV channel knob !

So if they were all jumbled up together on the same dial and you

wanted to tune your radio from one AM station to another, you

might have to crank through enough space for 600 radio stations ...

or even 1200 or 1800 of them ... to go to the AM signal you want.

And maybe even worse than that ! I'm sure you've never heard what

a TV signal SOUNDS like on a radio. It is horrendous, and it is loud !

It sounds like a thousand cats shrieking at each other, and it never stops.

That's another good reason to move the TV transmissions to frequencies

where radios will never hear them. If radios just randomly tuned in to a

TV picture signal every now and then, a lot of people would be shocked

out of their socks. They would stop listening to radio, and thousands of

advertisers would not like that.

For the second question ...

OK, so we don't mix radio and TV in the same band of frequencies.

But why does each station need its own frequency ? Why not just

put every radio station on one frequency, and every TV station on

a single frequency that's different from the radio frequency ?

The answer is: It's because people don't want to listen to two radio

stations at the same time, or watch two TV movies at the same time.

We like to make our choice, and then watch them or listen to them

one at a time. And FREQUENCY is the only way our radios and TVs

know how to pick out ONE and ignore all the others.

If there are two, or 5 or 10 stations all on the same frequency within

10 or 20 miles from you, then when you tune your radio to that frequency,

you HEAR two, or 5 or 10, songs, church services, newscasts, political

speeches, or commercials, all at the same time.

So if all radio stations were on the same frequency, or all TV stations

were all on the same frequency, then any time you turned on your

radio or TV, you'd see or hear all of them together. Radio and TV

would completely lose their entertainment value, everybody would

give up watching and listening, and once again ... thousands of

advertisers would not like that.

After all, advertising is the main reason why we have so much radio and TV at all. The advertiser buys, the broadcaster sells, and YOUR eyes and ears are the product.

5 0

3 years ago

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