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Nataly [62]
3 years ago
15

All are examples of electric forces except _________. A. a neutron pushing on another neutron B. an electron pushing on a proton

C. an electron pushing on another electron
Physics
1 answer:
Contact [7]3 years ago
6 0

Answer:

A) a neutron pushing on another neutron.

Explanation:

Neutrons have no charge, therefore, there is no electric force among them. Protons and electrons on the other hand do have electric charge (electrons negative charge and protons positive charge) that generate electric forces between them that can be repelling forces (if the charges are of the same sign), or attractive forces (if the charges are of opposite signs).

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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
Explain a situation in which you can accelerate even though your speed doesn’t change.
Serga [27]
"Acceleration" does NOT mean speeding up.  It also doesn't mean
slowing down.  Acceleration means ANY change in the speed
OR DIRECTION of motion.

The only kind of motion that's NOT accelerated is motion at a steady
speed AND in a straight line.

Even when your speed is steady, you're accelerating if your direction
is changing.

A few examples:
(no speeds are changing):

-- driving on a curved road, or turning a corner
-- going around a curve on a skateboard, a bike, or a Segway
-- running on a quarter-mile track
-- an Indy car cruising a practice lap around the track
-- water spinning, getting ready to go down the drain
-- any point on the blade of a fan
-- the little ball going around the inside of a Roulette wheel
-- the Moon in its orbit around the Earth
-- the Earth in its orbit around the sun
4 0
3 years ago
HELP ME PLZ WILL GIVE BRAINLIEST ALSO 24 POINTS
Ludmilka [50]

Answer:

The highest vertical position is where your maximum potential energy lies. At the highest altitude point of course ! This is when the kinetic energy is only due to horizontal motion (since the vertical component reaches zero).

Explanation:

i looked it up ok

6 0
3 years ago
Dave's sister gets very stressed out during final exam time; however, Dave isn't affected by the stress. What does this illustra
dmitriy555 [2]
The answer is b personal stress
6 0
2 years ago
Physical science is the study of
kakasveta [241]
Physical science is the study, measurement and observation of nonliving objects whereas biological.
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