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Ulleksa [173]
3 years ago
9

Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force

on one of the masses?
Physics
1 answer:
xeze [42]3 years ago
7 0

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
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A truck traveled 400 meters north in 80 seconds and then it traveled 300 meters east in 70 seconds. The magnitude of the average
Varvara68 [4.7K]

Answer:

3.3m/s

Explanation:

You first get the total time (80 + 70 = 150s).

Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.

d= 500m.

So now that you have displacement and time, you can find the velocity:

v=d/t

v=500/150

v=3.3

5 0
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Which of the following are examples of projectile motion?
denis-greek [22]

Answer:

A or C    I would pick C

Explanation:

8 0
3 years ago
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8 0
2 years ago
A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick
Rom4ik [11]
Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
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= 4.0 J
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5 0
2 years ago
A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0
3 years ago
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