1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alexira [117]
3 years ago
8

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

the distance between the particles is decreased to 1?
Physics
1 answer:
klemol [59]3 years ago
4 0

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

You might be interested in
a boat traveling at a speed of 1.63 m / s has a kinetic energy of 52000 J what is the mass of the boat
BARSIC [14]

Answer:

39,098‬ kg

Explanation:

KE = .5 * m * v^2

52000=.5m * 1.63^2

52000=.5m * 2.66

52000/2.66=.5m

19,549 = .5m

19549/.5 = m

m=39,098‬ kg

5 0
3 years ago
Does anyone know 19 and 20 ??
harina [27]
B I think for 19 and D for 20
4 0
3 years ago
Starting from equilibrium at point 0, what point on the pv diagram will describe the ideal gas after the following process? lock
Anna71 [15]
Since the product of P·Vis constant along an isotherm, an expansion to twice the volume implies a pressure reduction to half the original pressure. I hope my answer has come to your help. God bless and have a nice day ahead!<span>
</span>
3 0
3 years ago
Read 2 more answers
A sports car has an average acceleration of 5.81 m/s2. How long does it take for the car to reach 60.0 mi/h, if it starts from r
Aleks04 [339]

Answer:

  4.617 s

Explanation:

The speed of 60 mi/h can be converted to m/s:

  (60 mi/h) × (1609.344 m/mi) × (1 h)/(3600 s) = 26.8244 m/s

The relationship between speed and acceleration is ...

  v = at

  t = v/a = (26.8244 m/s)/(5.81 m/s²) ≈ 4.617 s

It will take the car 4.617 seconds to reach 60 mi/h starting from rest.

5 0
3 years ago
Read 2 more answers
A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N
Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

E_k = W = 40.2 j

mv^2/2 = 40.2

20v^2/2 = 40.2

10v^2 = 40.2

v^2 = 4.02

v = \sqrt{4.02} = 2 m/s  

3 0
3 years ago
Other questions:
  • ___________ light on the electromagnetic spectrum actually warms you up.
    12·1 answer
  • When would you use a compound microscope?
    9·1 answer
  • A 500-kg roller coaster car travels with some initial velocity along a track that is 5 m above the ground. The car goes down a s
    5·1 answer
  • A truck has a mass of 3000 kg and a velocity<br> of 10 m/s. Calculate momentum!
    13·2 answers
  • Which type of investigation is lised to compare two subiects?
    7·1 answer
  • The earth has about 80 times the mass of the earth's moon. The gravitational force exerted on the moon by the earth has what rel
    11·1 answer
  • How have workers in the past tried to better their lives? by forming "working schools" where they could continue and improve the
    5·1 answer
  • Write fulk form of FPS?<br>​
    13·1 answer
  • Draw a well labelled diagram of a refrigerator
    5·1 answer
  • if the spring balance with the iron ball is relaesed what reading will the spring balance record and wht
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!