The change in state of a liquid to a gas

which way would 2 negatively charged balloons naturally move? what would that do to the amount of potential energy stored in the

zheka24 [161]

Answer:

gsg

Explanation:

5 0

3 years ago

A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,

Marina CMI [18]

Answer:

B. Attract each other with a force of 10 newtons.

Explanation:

Statement is incorrectly written. The correct form is: A $+10\,\mu C$ charge and a $-10\,\mu C$ at a distance of 0.3 meters.

The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$ (1)

Where:

$F$ - Electrostatic force, in newtons.

$\kappa$ - Electrostatic constant, in newton-square meters per square coulomb.

$|q_{A}|,|q_{B}|$ - Magnitudes of electric charges, in coulombs.

$r$ - Distance between charges, in meters.

If we know that $\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C$ and $r = 0.3\,m$ , then the magnitude of the electrostatic force is:

$F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}$

$F = 9.987\,N$

In consequence, correct answer is B.

4 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

melisa1 [442]

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

3 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g

Vlada [557]

Answer:

(a) L = 128.75 m

(b) L = 182.56 m

(c) L = 114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

L = 128.75 m

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

L = 182.56 m

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

L = 114.28 m

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

Mass of Gold = 7.68 kg = 7680 gram

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

Cost of Gold Wire = $ 307040

7 0

3 years ago

The change in state of a liquid to a gas​

The change in state of a liquid to a gas