Which correctly lists three places that fresh water is found?

Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the

allsm [11]

Answer:

r = 6.5*10^-3 m

Explanation:

I'm assuming you meant to ask the diameters of the disk, if so, here's it

Given

Quantity of charge on electron, Q = 1.4*10^9

Electric field strength, e = 1.9*10^5

q = Q * 1.6*10^-19

q = 2.24*10^-10

E = q/ε(0)A, making A the subject of formula, we have

A = q / [E * ε(0)], where

ε(0) = 8.85*10^-12

A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)

A = 2.24*10^-10 / 1.6815*10^-6

A = 1.33*10^-4 m²

Remember A = πr²

1.33*10^-4 = 3.142 * r²

r² = 1.33*10^-4 / 3.142

r² = 4.23*10^-5

r = 6.5*10^-3 m

3 0

2 years ago

Can someone please help me

Tamiku [17]

<em>F</em> = 153 N

$\theta = 11.3°$

Explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components $F_{x}$ and $F_{y}$ :

$F_{x}$ = 350 N - 200 N = 150 N

$F_{y}$ = 180 N - 150 N = 30 N

The magnitude of the resultant force <em>F</em> is given by

$F = \sqrt{F_{x}^{2}+F_{y}^{2}}$

$\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}$

$\:\:\:\:\:\:=153\:N$

To find the direction $\theta$ , we use

$\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2$

or

$\theta = \tan^{-1}(0.2) = 11.3°$

4 0

2 years ago

Your destroyer has a RADAR antenna height of 40 m. Using RADAR, what is the maximum detection range of a patrol boat with a mast

Lisa [10]

Answer:

The maximum detection range is 39.75 km

Explanation:

Given that;

Antenna height h1 = 40 m

Target height ( patrol boat mast ) h2 = 11 m

Using RADAR, what is the maximum detection range = ?

Using RADAR

we know that; Maximum detection range = (√17h1 + √17h2) km

where h1 and h2 are heights of the antenna and target height in meters

so we substitute in our values

Maximum detection range = (√(17 × 40) + √(17 × 11)) km

Maximum detection range = (√680 + √187) km

Maximum detection range = (26.0768 + 13.6747) km

Maximum detection range = 39.75 km

Therefore, The maximum detection range is 39.75 km

6 0

3 years ago

What is the potential energy of a 2.3g penny that is held at the top of the cn tower (553m)?

SpyIntel [72]

Answer:

12.46 joules

Explanation:

The formula of potential energy is:

$\displaystyle{E_p = mgh}$

where m is mass, g is gravitational force (9.8 m/s^2) and h is height.

Mass is given to be 2.3 grams.
Gravitational Force is defined as 9.8 m/s^2.
Height is given as 553 meters.

However, mass must be in kilograms (kg) unit so we will be converting the gram to kilogram:

We know that a gram equals to 0.001 kg so 2.3 grams will equal to 2.3 * 0.001 = 0.0023 kg.

Therefore, substitute known information in the formula of potential energy:

$\displaystyle{E_p = 0.0023 \times 9.8 \times 553}\\\\\displaystyle{E_p = 12.46 \ \, \sf{Joules}}$

Therefore, the potential energy is 12.46 joules.

6 0

1 year ago

Two tiny, spherical water drops, with identical charges of -4.29 Ã 10-16 C, have a center-to-center separation of 0.845 cm. A) W

RSB [31]

Answer:

Electrostatic force, $F=2.31\times 10^{-17}\ N$

Number of electrons, n = 2681 electrons

Explanation:

Given that,

Charges, $q_=q_2=-4.29\times 10^{-16}\ C$

Separation between charges, $r=0.845\ cm=0.845\times 10^{-2}\ m$

(a) Let F is the magnitude of the electrostatic force acting between them. The electric force between charges is given by :

$F=\dfrac{kq^2}{r^2}$

$F=\dfrac{9\times 10^9\times (4.29\times 10^{-16})^2}{(0.845\times 10^{-2})^2}$

$F=2.31\times 10^{-17}\ N$

(b) Let n be the excess electrons on each drop, giving it its charge imbalance. It can be calculated using quantization of electric charge as :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{4.29\times 10^{-16}}{1.6\times 10^{-19}}$

n = 2681.25 electrons

or

n = 2681 electrons

So, the number of excess electrons on each drop is 2681 electrons. Hence, this is the required solution.

4 0

3 years ago