the mass of one water drop is 0.0008kg and the gravitational field strength is 10N/kg what is its weight

Take one jar with salt water and another with pure water .put an egg into both jars. In which the jar egg floated over the liqui

DanielleElmas [232]

Answer:

if the which coduct salt has more density so the egg floats .

Explanation:

8 0

3 years ago

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

A boat floats in water a. Archimedes b. Bernoulli c. Density d. Pascal e. Pressure

Ivenika [448]

The answer is Density !, Do you also need an example ?

Rate this the brainliest answer , Thank youuu !

7 0

3 years ago

Read 2 more answers

Electric pressure is measured is a, current .b,volts,c, Walt's,d, Ampere,e, force

kolbaska11 [484]

｡☆✼★ ━━━━━━━━━━━━━━ ☾

The correct option would be B. volts

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

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5 0

4 years ago

On a smooth horizontal floor, an object slides into a spring which is attached to another mass that is initially stationary. Whe

saul85 [17]

Answer: (E) Momentum and mechanical energy

Explanation:

The momentum and the mechanical energy is basically conserved during the given interaction process as the forces on the given system are in the form of internal nature and then the momentum are get conserved.

According to the given question, on the smooth floor when an object are slides by using the spring then the momentum and the mechanical energy are conserved.

The mechanical energy is the combination of both the kinetic and the potential energy that is used for doing some amount of work. Therefore, Option (E) is correct answer.

5 0

3 years ago