All categories

3 years ago

The formula K2S indicates that

Physics

1 answer:

jekas [21]3 years ago

4 0

There are two atoms of potassium bonded to one atom of sulfur.

You might be interested in

What are the starting types of energy and what are the ending types of energy?

vladimir2022 [97]

Answer: These are the starting of than energy i think chemical, electrical, radiant, mechanical, thermal

6 0

2 years ago

(a) How much gravitational potential energy (relative to the ground on which it is built) is stored in an Egyptian pyramid, give

Rom4ik [11]

Answer:

$2.66832\times10^{12}$

2,668,320,000

Explanation:

Potential energy is given as the product of mass, acceleration due to gravity and height.

PE=mgh

Where m is the mass of pyramid, g is acceleration due to gravity and h is height at any point.

In reference to the earth, we substitute h for 34.0 m and g with 9.81 m/s2 then taking m as 8000000000 kg we get that

PE=mgh=8000000*9.81*34.0=2,668,320,000 J

In scientific notafion, this is expressed as $2.66832\times10^{12}$

3 0

3 years ago

Experimental Inquiry: What factors influence the loss of nutrients from a forest ecosystem?

maxonik [38]

Many factors including Decomposition, energy conversion, littering, erosion, etc.

6 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

Irina18 [472]

Answer:

a. $q_C$ experiences the greatest net force and $q_B$ experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

$\displaystyle F=K\frac{q_1q_2}{d^2}$

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

$F_{TA}=-F_B-F_C+F_D$

Computing the individual forces we have

$\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}$

$\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}$

$\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}$

The total force on qA is:

$\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}$

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

$\displaystyle F_{TB}=F_A-F_C+F_D$

$\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}$

$\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

$\displaystyle F_{TC}=F_A+F_B+F_D$

$\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}$

$\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

$\displaystyle F_{TD}=-F_A-F_B-F_C$

$\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}$

$\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}$

Comparing the relative values of all the forces

$\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|$

This means that qc experiences the greatest net force and qB experiences the smallest net force

The ratio of the greatest to the smallest forces is

$\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9$

5 0

3 years ago

A boy im50kg at rest on a skateboard is pushed by another boy who exerts a force of 200 N on him. If the first boy's

sergey [27]

Answer:

Time, t = 2 seconds

Explanation:

Given the following data;

Mass, m = 50 kg

Initial velocity, u = 0 m/s (since it's starting from rest).

Final velocity, v = 8 m/s

Force, F = 200 N

To find the time, we would use the following formula;

$F = \frac {m(v - u)}{t}$

Making time, t the subject of formula, we have;

$t = \frac {m(v - u)}{F}$

Substituting into the formula, we have;

$t = \frac {50(8 - 0)}{200}$

$t = \frac {50*(8)}{200}$

$t = \frac {400}{200}$

Time, t = 2 seconds

5 0

3 years ago