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Rudiy27
3 years ago
5

how can you tell, as you walk close to a parked car, if it had been running recently? describe your reasoning in terms of energy

flows.
Physics
1 answer:
Blababa [14]3 years ago
4 0
This question is probably referring to heat energy transferring from the car to its surroundings.
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Which of these pairs of atoms are isotpoes? (Physical Science) Pair A Pair B Pair C # protons 6 8 5 2 12 12 # neutrons 8 8 5 3 1
Aleksandr-060686 [28]

Answer:

I guess that the atoms are:

Protons:     6     8    5    2    12    12

Neutrons:  8     8     5   3     13    14

Now, two atoms are isotopes if they share the same number of protons (so both atoms are the same element) but they have a different number of neutrons.

From the given options, the only two that have the same number of protons but a different number of neutrons are:

Protons 12, neutrons 13

and

Protons 12, neutrons 14.

These two are isiotopes.

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3 years ago
Is it true that playing badmenton help you to become a better person?
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3 years ago
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Explain why the surface of Venus is hotter than the surface of Mercury
Hitman42 [59]
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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
How many newtons of force are represented by the following amount: 3 kg·m/sec^2?
Jet001 [13]
1N=1kg•m/s^2 so the answer is 3N
8 0
3 years ago
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