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
777dan777 [17]
3 years ago
12

A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

he x-axis at −8 cm . −10−8 −6 −4 −2 2 4 6 8 10 8 µ C 3 µ C − 3 µ C x → (cm) What is the magnitude of the total electrostatic force on a third particle with charge −3 µC placed on the x-axis at 2 cm
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

You might be interested in
A constant force of 11.8 N in the positive x direction acts on a 4.7-kg object as it moves from the origin to the point (1.6i –
zhenek [66]

Answer:

W = 18.88 J

Explanation:

Given that,

Constant force, F = 11.8 N (in +x direction)

Mass of an object, m = 4.7 kg

The object moves from the origin to the point (1.6i – 4.6j) m

We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :

W=F{\cdot} d\\\\W=(11.8i){\cdot} (1.6i-4.6j)\\\\=11.8\times 1.6\\\\=18.88\ J

So, the work done by the given force is 18.88 J.

5 0
3 years ago
If there is an object in top of another object, why does the upper object exert a downward normal force?​
rusak2 [61]

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.  

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

4 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
4 years ago
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas
Arada [10]

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

7 0
3 years ago
How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

3 = (40 - 10) / t          Multiply both sides by t

3*t = t(40 - 10)/t        Combine. Cancel t's on the right

3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

6 0
2 years ago
Other questions:
  • A force of 50 pounds is directed 30 from the horizontal, find the vector that represents this force.
    12·1 answer
  • Collisions between galaxies are thought to:
    6·1 answer
  • Which phase of matter has the MOST Kinetic Energy?
    10·2 answers
  • At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large, compressed spring. Th
    15·2 answers
  • What is the low end of the range of surface temperature of white stars
    7·1 answer
  • What does a measured number tell you?
    12·1 answer
  • A man of mass 50kg ascends a flight of stairs 5m high in 5seconds. If acceleration due to gravity is 10ms-2 the power expended i
    10·2 answers
  • Which part of the iceberg displaces water equal in weight to the buoyant force
    6·2 answers
  • How much work is done by a crane that lowers 1000N of material a distance of 150
    11·2 answers
  • The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of th
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!