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snow_tiger [21]
3 years ago
10

What's the Coulomb's law?

Physics
2 answers:
Ulleksa [173]3 years ago
4 0

<span>
In layman's term: </span>like charges don't attract while opposite charges do<span>electrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. </span><span>there is also something else about this  law that I don't quite remember.</span>

<span>___________________________________________________</span>

<span />Here is the formula:

<span>F = k x Q1 x Q2/d^<span>2</span></span>

<span>What the formula means:</span>

F=force between charges

Q1 and Q2= amount of charge

d=distance between these two charges

k= Coulombs constant (proportionally constant)

________________________________________________

I think that about covers it and hopefully this helped.

nexus9112 [7]3 years ago
3 0
Coulomb's law states that if you place two particles of respective charges q_1,q_2 at a distance d from each other, one will exert the following force on the other :

\vec{F}=k\frac{q_1q_2}{d^2}\vec{u} where \vec{u} is a unit vector from the first charge to the other and k is a positive constant.

A direct consequence of this is that two charges of same sign repel each other, while two charges of opposite signs attract each other.

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3 0
3 years ago
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Ahmed fills a basket with shopping weighing 10 kg. How much work is being done on the shopping basket when he lifts it verticall
NISA [10]

The work done by the shopping basket is 147 J.

<h3>When is work said to be done?</h3>

Work is said to be done whenever a force moves an object through a certain distance.

The amount of work done on the shopping basket can be calculated using the formula below.

Formula:

  • W = mgh

Where:

  • W = Amount of work done by the basket
  • m = mass of the shopping basket
  • h = height of the shopping basket
  • g = acceleration due to gravity.

Form the question,

Given:

  • m = 10 kg
  • h = 1.5 m
  • g = 9.8 m/s²

Substitute these values into equation 2

  • W = 10(1.5)(9.8)
  • W = 147 J.

Hence, The work done by the shopping basket is 147 J.

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6 0
2 years ago
A ball is projected at an immovable wall with a speed vi and bounces back the wall in such a manner that it only has 1/3 of its
sergey [27]

The fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

The given parameters;

  • <em>initial speed of the ball, = vi</em>
  • <em>final momentum of the ball, Pf = ¹/₃Pi</em>

The initial and final momentum of the ball is calculated as;

P_i = m_ivi

P_f = m_fv_f = \frac{1}{3} m_iv_i

The initial and final kinetic energy of the ball is calculated as;

K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i

The change in the kinetic energy is calculated as;

\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i

Thus, the fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

Learn more here:brainly.com/question/18566218

5 0
3 years ago
a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th
anzhelika [568]

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

F_{1} = 250

A_{1} = Area of the small piston = πr^{2}

A_{1} = 22/7 x 0.4^{2}

A_{1} = 0.5 m^{2}

F_{2} = ?

A_{2} = Area of the large piston = πr^{2}

A_{2} = π x 1

A_{2} = 3.14 m^{2}

To calculate the force on the large piston, we will use the below formula

F_{1}/ A_{1} = F_{2} / A_{2}

Substitute all the parameters into the equation

250/0.5 =  F_{2}/3.14

F_{2} = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

F_{1}/ 0.25A_{1} = F_{2} / A_{2}h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

Learn more here: brainly.com/question/13596980

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2 years ago
Hey swimmer swim 50 m in 20 seconds what is their average velocity
fenix001 [56]
It’s 2.6 for the average velocity
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3 years ago
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