The electrostatic forces between two charges are
F = (9 x 10⁹) x (charge-1) x (charge-2) / (distance between them)² .
The charges in this question are each (1 x 10⁻⁶ Coulomb).
Force = (9 x 10⁹) (1 x 10⁻⁶) (1 x 10⁻⁶) / (4.0)
= (9 x 10⁹) (1 x 10⁻¹²) / (4.0)
= 0.00225 Newton .
-- The force on each mass is the same.
Each mass is acted on by a force of 0.00225 newton,
directed either toward the other one or away from it.
-- If the charges both have the same sign, then the forces
are repulsive ... driving the masses apart.
-- If the charges have different signs, then the forces
are attractive ... drawing the masses together.
-- For as far as you've gone in copying this question ... asking
only for the forces between the charged objects ... you don't
need to know their masses, or about friction with the table, or
any of that stuff. None of that has any effect on the forces
between the charged masses.
You'll need that information when you get down to the next part
of the question, where it's probably going to ask you for the
acceleration of each mass due to the electrostatic force on it.
You know the force on each mass now. All you have to remember
is F = (mass) x (acceleration) , and grind it out.
Don't you dare post the next part on Brainly ! You have everything
you need to do it yourself now, and you'll learn a lot more that way.
Answer:
The force that you must exert on the balloon is 1.96 N
Explanation:
Given;
height of water, h = 4.00 cm = 4 x 10⁻² m
effective area, A = 50.0 cm² = 50 x 10⁻⁴ m²
density of water, ρ = 1 x 10³ kg/m³
Gauge pressure of the balloon is calculated as;
P = ρgh
where;
ρ is density of water
g is acceleration due to gravity
h is height of water
P = 1 x 10³ x 9.8 x 4 x 10⁻²
P = 392 N/m²
The force exerted on the balloon is calculated as;
F = PA
where;
P is pressure of the balloon
A is the effective area
F = 392 x 50 x 10⁻⁴
F = 1.96 N
Therefore, the force that you must exert on the balloon is 1.96 N
Answer:
A
Explanation:
a statement that can be tested through the scientific method
Answer:
i = 0.00077A
Explanation:
Given:
loop radius, r = 3.0 cm = 0.03 m
Area, A = π x r² = π x 0.03² = 0.0028 m²
Magnetic Field, B = 0.75 T
Loop resistance, R = 18 Ω
time, t = 0.15 seconds
Now,
the induced emf is given as:
EMF = - BA/t .......1
Likewise,
EMF = iR.......2
Equate 1 and 2
iR = - BA/t
i = - BA/tR
i = 0.75×0.0028/0.15×18
i = 0.0021/2.7
i = 0.00077A
