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Dennis_Churaev [7]
3 years ago
14

20 POINT QUESTION

Physics
2 answers:
Valentin [98]3 years ago
8 0
Label A: sublimation
Label B: condensation
Label C: melting
PIT_PIT [208]3 years ago
4 0

Label A: sublimation

Label B: condensation

Label C: melting

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sergey [27]

Explanation:

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UPVOTES FOR EVERY ANSWER!!!!!!!!
Goshia [24]
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A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis
lawyer [7]

Answer:

The value of the electric field is E_{net} = \dfrac{r \textbf{b}}{2\epsilon_{0}}

Explanation:

We know that the electric field inside a solid cylinder at a distance \textbf{r} from the centre is given by

E = \dfrac{\rho \textbf{r}}{2 \epsilon_{0}}

Let's consider the cross-section of the cylinder as shown in the figure. Let `O' be the centre of the long solid insulating cylinder having radius 'R'. Also consider that O' be the cetre of the hole of radius 'a' situated at a distance 'b' from 'O'. Given, the volume charge density of the material is 'r'. So, the volume charge density inside the hole will be '-r'. Let's consider 'P' be any arbitrary point inside the hole situated at a distance 's' from O'.

So, the electric field 'E_{O}' due to the long cylinder at point 'P' is given by

E_{O} = \dfrac{r \textbf{c}}{2 \epsilon_{0}}

and the electric field 'E_{O'}'due to the hole at point 'P' is given by

E_{O'} = \dfrac{\rho \textbf{s}}{2 \epsilon_{0}}

So the net electric field (E_{net}) inside the hole is given by

E_{net} = E_{O} - E_{O'} = \dfrac{r}{2\epsilon_{0}}(\textbf{c - s}) = \dfrac{r \textbf{b}}{2\epsilon_{0}}

5 0
4 years ago
Learning good work ethics will help you develop _____.
Pavel [41]

Answer:

a sense of industry

Explanation:

I just had this question and got it right

3 0
3 years ago
A 5000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses 0.1 meters. Assume no damping. a) Find the spring
ludmilkaskok [199]

Answer:

k = 0.5 MN/m

Explanation:

Mass of the railcar, m = 5000 kg

Speed of the rail car, v = 1 m/s

The Kinetic energy(KE) of the railcar is given by the equation:

KE = 0.5 mv²

KE = 0.5 * 5000 * 1²

KE = 2500 J

The spring's compression, x = 0.1 m

The potential energy(PE) stored in the spring is given by the equation:

PE = 0.5kx²

PE = 0.5 * k * 0.1²

PE = 0.005k

According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring

KE = PE

2500 = 0.005k

k = 2500/0.005

k = 500000 N/m

k = 0.5 MN/m

8 0
4 years ago
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