How many outer-orbital electrons are found in an atom of:a) Na?b) P?c) Br?d) I?e) Te?f) Sr?

Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot

guapka [62]

Answer:

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

$E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}$

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

$R = \frac{d}{\sqrt2}$

now we have

$E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}$

so the force on the charge is given as

$F = QE$

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

3 0

3 years ago

The universal force that is effective over the longest distance between very massive objects like planets is

fenix001 [56]

The universal force that is effective over the longest distance between very massive objects like planets is Gravity

8 0

3 years ago

What are the limitations of using solar panels?

Anit [1.1K]

Answer:

The amount of the sun energy that could be collected.

Explanation:

Some limitations are, the amount of the sun's energy that could be collected the radiation of sun is nearly fixed. The place where we can put the solar panels are also limited.

7 0

3 years ago

Please help me on this!

aalyn [17]

Answer:

erosion

Explanation:

grows into the eirth each year. the atmosphere and how the worldchanges every year

5 0

2 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago