All categories

3 years ago

Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?

1 answer:

5 0

The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = (k * q1 * q2) / (r^2) where:
q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9 N. m2/ C2
r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9 x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n

You might be interested in

What is the final step in the fourth stage of technological design

Ierofanga [76]

Answer:

after a product has been improved and approved? reporting the results finding ways to lower costs selling a prototype determining criteria.

Explanation:

5 0

2 years ago

A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

= $\frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}$

= 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0

2 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Kitty [74]

Answer:

Q = 282,000 J

Explanation:

Given that,

The mass of liquid water, m = 125 g

Temperature, T = 100°C

The latent heat of vaporization, Hv = 2258 J/g.

We need to find the amount of heat needed to vaporize 125 g of liquid water. We can find it as follows :

$Q=mH_v\\\\Q=125\ g\times 2285\ J/g\\\\Q=282250\ J$

Q = 282,000 J

So, the required heat is 282,000 J .

6 0

2 years ago

If you run 10 km east and then turn around and run 7 km west, how much total distance have you travelled? A. 17 km B. 70 km C. 3

Tom [10]

It's A 17 km because the total km walked is 17.

4 0

2 years ago

Read 2 more answers

imagine that you are on a spaceship that is far from the sun or any planet. Explain why a 500 kg piece of equipment would be dif

Mila [183]

Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.

8 0

2 years ago