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Oksana_A [137]
2 years ago
8

A block of mass 8 m can move without friction

Physics
1 answer:
nekit [7.7K]2 years ago
5 0

Answer:

Let M1 = 8 kg and M2 = 34 kg

F = M a = (M1 + M2) a

F = M2 g     the net force accelerating the system

M2 g = (M1 + M2) a

a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2

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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl
Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

U = \frac{1}{2}\epsilon_0 E^2

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}

Creating a Gaussian surface being the lateral surface area of a cylinder:
A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}

Now, we can calculate the energy density using the equation:
U = \frac{1}{2} \epsilon_0 E^2

Plug in the expression for the electric field and solve.

U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
V = \pi r^2L \\\\dV = 2\pi rLdr

Now, we can take the integral of the above expression. Let:
r_i = inner cylinder radius

r_o = outer cylindrical shell inner radius

Total energy-field energy:

U = \int\limits^{r_o}_{r_i} {U_D} \, dV =   \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr

Plug in the equation for the electric field energy density and solve.

U =   \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\

Bring constants in front and integrate. Recall the following integration rule:
\int {\frac{1}{x}} \, dx  = ln(x) + C

Now, we can solve!

U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}

And here's our equation!

3 0
2 years ago
Radon-222 transforms into polonium-218 and an alpha particle (he-4). what else is created?
ELEN [110]
Hm pretty much that's all for alpha decay... But always after any alpha or beta decay, gamma photons are released because the daughter products end up in an excited state so they have to release energy through gamma decay
6 0
3 years ago
Why do yall not hate us country people?
sveticcg [70]

Answer:

Because we don't?

Explanation:

4 0
3 years ago
What is the mass of an object going 25 m/s and has a momentum of 140<br> kg m/s?
slega [8]

Answer:

The answer would be 5.6 kg of a mass.

Explanation:

p= mv

m= p/ v

The explanation of that is:

momentum = mass× velocity

mass= momentum / velocity

Hope this helped!

7 0
3 years ago
The specific heat of water in its solid phase (ice) is 2090 J/(kg K), while in the liquid phase (water) its specific heat is 419
oee [108]

Answer:

d. 149 ⁰C.

Explanation:

Given;

mass of the block of ice, m = 2 kg

specific heat capacity of the ice, C = 2090 J/(kgK)

initial temperature of the ice, t₁ = -90 ⁰C

heat added to the ice, H = 1,000,000 J

let the final temperature of the ice = t₂

The final temperature of the ice after adding the heat is calculated as follows;

H = mC_{ice} \Delta t\\\\H = mC_{ice}  (t_2 - t_1)\\\\1,000,000 = 2 \times 2090 \times (t_2 - (-90))\\\\1,000,000 = 4,180(t_2 +90)\\\\1,000,000 = 4,180t_2 + 376,200\\\\1,000,000 - 376,200 = 4,180t_2\\\\623,800 = 4,180 t_2\\\\t_2 = \frac{623,800}{4,180} \\\\t_2 = 149 \ ^0C

Therefore, the new temperature of the water is 149 ⁰C.

4 0
3 years ago
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