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Vikki [24]
3 years ago
7

A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam

e charge. this lower sphere is fixed and cannot move. part a what is the magnitude of the electric force between the spheres?
Physics
1 answer:
Allushta [10]3 years ago
4 0
For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹ 
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
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A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement.
zalisa [80]

Explanation:

EE = ½ kx²

EE = ½ (800 N/m) (6 m)²

EE = 14,400 J

8 0
3 years ago
Will give brainliest!!
givi [52]
A. Move the candle to the right, or the focal point to the left.

For a convex lens, the closer an object is to the focal point, the larger it’s image is (and therefore the greater the magnification is). The two ways you could make the candle be closer to the focal point are to move the candle to the right, or the focal point to the left.
4 0
3 years ago
Gravity causes all falling objects to accelerate at a rate of 98 m/s2.<br> O True<br> O False
aleksandrvk [35]
It’s true the acceleration of falling objects on earth due to gravity is 98ms2
8 0
2 years ago
What is the distance fallen for a freely falling object 1 s after being dropped from a rest position? What is the distance for a
kondor19780726 [428]
<h2>Distance traveled in 1 second after drop is 4.9 m</h2><h2>Distance traveled in 4 seconds after drop is 78.4 m</h2>

Explanation:

We have s = ut + 0.5at²

For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²

Substituting

                 s = 0 x t + 0.5 x 9.8 x t²

                 s = 4.9t²

We need to find distance traveled in 1 s and 4 s

Distance traveled in 1 second

                   s = 4.9 x 1² = 4.9 m

Distance traveled in 4 seconds

                   s = 4.9 x 4² = 78.4 m

Distance traveled in 1 second after drop = 4.9 m

Distance traveled in 4 seconds after drop = 78.4 m

5 0
2 years ago
In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it
anastassius [24]

Answer: 3.48g

Explanation:

here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

Velocity of blood = 56.5cm = 0.565m

mass of blood * 0.565 = 54kg * (0.000063/0.160)

mass of blood * 0.565 = 54 * 0.00039375

mass of blood * 0.565 = 0.001969

mass of blood = 0.00348kg

Thus, the mass of blood that leaves the heart is 3.48g

7 0
3 years ago
Read 2 more answers
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