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3 years ago

What is thE Rquation fo calculating the electrical force between two charges?

2 answers:

8 0

The direction of the electrical force is dependent upon whether the charged objects are charged with like charge or opposite charge and upon their spatial orientation. By knowing the type of charge on the two objects, the direction of the force on either one of them can be determined with a little reasoning.

umka21 [38]3 years ago

4 0

The equation for calculating the electrical force between the two charges is given by Coulomb's law

F=\frac{k q1 q2}{r^{2}}

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Greater is the magnitude of the charges, greater will be the force between the two charges. The force is inversely proportional to the square of the distance between the charges.

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What was one main point of Dalton's atomic theory

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All matter is composed of atoms, indestructible building blocks.

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4 years ago

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If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s

Vlad1618 [11]

The magnitude of the air drag is 784 N

Explanation:

An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:

$F_D = mg$

where

$F_D$ is the magnitude of the air drag

m is the mass of the object

g is the acceleration of gravity

In this problem, we have

m = 80 kg is the mass of the airman

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting into the formula, we find:

$F_D = (80)(9.8)=784 N$

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3 years ago

An industrial overhead door has a chain and sprocket system designed to reduce the force needed by an operator to open and close

mylen [45]

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4 years ago

A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.