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valentina_108 [34]

3 years ago

15

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t

hem to be 83.4 N?

2 answers:

nataly862011 [7]3 years ago

7 0

F = q₁*q₂* C / r²

F =83.4
q₁ = q₂ = 3.2 * 10⁻⁹
C couloumb constant 8.9 * 10⁹

solve for r

STatiana [176]3 years ago

4 0

I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!

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The Earth’s gravitational force that attracts the sun is __________ the sun’s gravitational force to attract the Earth?

Ede4ka [16]

Hello,

Here is your answer:

The proper answer to this question is option D "<span>centrifugal".</span>

The force centrifugal is the force that holds the moon with the Earth and the centrifugal force is two times stronger with the Earth and the sun.

Your answer is D.

If you need anymore help feel free to ask me!'

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5 0

3 years ago

How much energy is required to change a 39.0-g ice cube from ice at -14.0°c to steam at 108°c?

Ann [662]

Heat is added to the mass as follows:
Q_-14-0 = mC_iΔT = 39*2.06*(0--14) = 39*2.06*14 = 1124.76 J
Q_0 = mC_f = 39*334 = 13026 J
Q_0-100 = mC_wΔT = 39*4.18*100 = 16302 J
Q_100 = mC_v = 39*2230 = 86970 J
Q_100-108 = mC_sΔT = 39*2.03*(108-100) = 39*2.03*8 = 633.36J

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7 0

3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

23. Why would an astronomical unit not be<br> used to measure distances on Earth?

Rashid [163]

One Astronomical Unit is the distance between the Earth and the Sun, this acts as a sort of a cosmic metre stick. Astronomers don't and cannot measure distances. Distances are merely inferred from what actually has been measured, such as an angle, a relative luminosity, a time period

3 0

3 years ago

A train leaves the station heading south on the tracks. It takes 5 seconds to reach 50 miles per hour. It completes the entire 1

Musya8 [376]

Well, in order to figure out the answer is to divide until you figure out how many miles they went per second. If it takes 5 seconds to reach 50 miles per hour it took 10 seconds per every 10 miles meaning each mile took 1 second. (Not actually possible but the answer) So, If it finished a 100 mile trip in 2 hours it took an hour for 50 miles. If it took 1 hour for 50 miles divide 60/50 which gets you 1.2 so it took 1.2 miles per minute meaning the car went 120 miles per hour I believe. I hope this helps :)

8 0

3 years ago

Read 2 more answers