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AVprozaik [17]
3 years ago
14

Explain your results: How can Earth’s gravity affect the water when the water isn’t actually touching the Earth? (2 pts) Do you

think the Moon’s gravity is affecting the water in the beaker? Why or why not? (2 pts) If you were on the Moon, do you think you would feel the pull of Earth’s gravity? Explain! (2 pts)
Physics
1 answer:
Levart [38]3 years ago
5 0

Answer:

Because the Earth has so much gravity, it can hold water, land, and life in it's atmosphere.

(Not sure what beaker you are talking about, so sorry) But I don't think the moon's gravity would have an effect on a beaker of water because the Earth's gravity is much more than the moon's.

I think you would be able to feel a little bit of Earth's gravity  on the moon because the Earth's gravity pulled the moon into orbit, therefore, gravity on Earth my have some effect on the moon.

hope this helps!

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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

7 0
3 years ago
Which units are used to express sound intensity?
grigory [225]

Answer: decibels

Explanation:

4 0
3 years ago
A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth
yulyashka [42]

Answer:

Explanation:

As we know the , equation of time period for simple pendulum ,

T = 2*pi*\sqrt{l/g}

hence putting values we get ,

the solution is in picture ,

please  

Brain-list it or support me at my U-Tube channel " ZK SOFT&GAMING " I will be thankful

5 0
3 years ago
Jared has a nut sack that weighs 14 ounces when weighed. If Becky puts an amount of force on them, x , what is the equation that
topjm [15]
X+14 bc that’s what it will equal.
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If two normal distributions each have the same mean, but one has a standard deviation of 1 and the other has a standard deviatio
likoan [24]

In comparison to a distribution with a standard deviation of 5, one has a curve with a higher peak. The variability will be greater the larger the standard deviation. It denotes increased variability in a distribution with a standard deviation of 5.

<h3>What do you mean by the term standard deviation?</h3>

The term "standard deviation" (or "") refers to a measurement of the data's dispersion from the mean. A low standard deviation implies that the data are grouped around the mean, whereas a large standard deviation shows that the data are more dispersed. In contrast, a high or low standard deviation indicates that the data points are, respectively, above or below the mean. A standard deviation that is close to zero implies that the data points are close to the mean. the curve at the top is more dispersed and has a greater standard deviation than the curve at the bottom, which is more concentrated around the mean and has a lower standard deviation.

To learn more about standard deviation, Visit:

brainly.com/question/14650840

#SPJ4

8 0
1 year ago
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