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den301095 [7]
2 years ago
7

In the fishbowl, the glass, water, rocks, and plastic plants are in thermal equilibrium. This situation means the temperature of

the rocks is...​
Physics
1 answer:
Assoli18 [71]2 years ago
7 0
To cold which it throws off the equilibrium of the other things and sticks to it longer
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Explain why ice floats in water. Explain why ice floats in water. Steam is more dense than water. Ice is the same density as wat
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Hey how's your day going

I hope after I answer that you understand and don't just paste my answer into your assignment!!! (<- read!!!)

Answer \|/

             

Ice is less dense than water.

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When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats

3 0
2 years ago
⦁ Match the following terms:
ki77a [65]

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

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4 0
3 years ago
Another name for a pivot point is the:
Makovka662 [10]

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I think it's c .... fulcrum

7 0
3 years ago
Question is attatched!
lina2011 [118]
All it does is lets him pull in a more convenient direction to raise the load. It has no effect on the required force.
4 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
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