The Tacoma narrows bridge collapsed due to?

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

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

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago

A water rocket uses an amount of water and pressurized air to send a plastic rocket several feet into the air. As the water and

Ilya [14]

Answer:

D.

For every action there is an equal and opposite reaction.

Explanation:

im doing the same one lol

6 0

3 years ago

The coefficients of static and kinetic frictions for plastic on wood are 0.50 and 0.40,respectively. How much horizontal force w

IrinaK [193]

The horizontal force needed to start the calculator moving from rest is 1.5 N

What is Kinetic friction?

It is defined as a force that acts between moving surfaces.

The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.

Here,

weight of calculator, N = 3 N

The coefficients of static frictions, µ (static) = 0.50

The coefficients of kinetic frictions, µ (kinetic) = 0.40

Now,

The horizontal force required = The static friction force

F = µ (static) * weight of calculator

F = 0.50 * 3.0

F = 1.5 N

Hence,

The horizontal force needed to start the calculator moving from rest is 1.5 N

Learn more about horizontal force here:

<u>brainly.com/question/21481680</u>

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#SPJ4

5 0

2 years ago

If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo

Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0

3 years ago

A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10

Alchen [17]