What is the regular rising and falling of ocean water due to the gravitational pull of the sun and moon?

HOLA, NECESITO AYUDA!

givi [52]

The electrostatic force between the two ions is $2.9\cdot 10^{-10} N$

Explanation:

The electrostatic force between two charged particle is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, the ion of sodium has a charge of

$q_1 = +e = +1.6\cdot 10^{-19} C$

while the ion of chlorine has a charge of

$q_2 = -e = -1.6\cdot 10^{-19}C$

And the distance between the two ions is

$r=282 pm = 282\cdot 10^{-12} m$

Substituting, we find the electrostatic force between the two ions:

$F=(8.99\cdot 10^9) \frac{(1.6\cdot 10^{-19})(-1.6\cdot 10^{-19})}{(282\cdot 10^{-12})^2}=-2.9\cdot 10^{-10} N$

where the negative sign simply means that the force is attractive, since the two ions have opposite charge.

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.

fredd [130]

Energy to lift something =

(mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm = 5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

= 2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$