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Irina-Kira [14]
3 years ago
12

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. How much wor

k is required to bring an additional electron from infinity to the origin?
Physics
1 answer:
mestny [16]3 years ago
3 0
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{e q}{r^2}

So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
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An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
4 years ago
A 360-N child is in a swing that is attached to a pair of ropes 1.90 m long. Find the gravitational potential energy of the chil
Igoryamba

Answer:

684J

Explanation:

So basically the formula for gravitational potential energy is Mass X Gravity X height. That is G.p.e = mgh

We don't have the mass but since we have the height, we multiply directly with the height since the quantity of weight is already given.

so G.p.e = 360 X 1.9 = 684J

Note that; The answer is in joules because g.p.e is work done.

Hope that was helpful!!

7 0
3 years ago
How much work is done if a force of 20N is used to move an object 6 metres? <br><br> pls help
Dmitriy789 [7]

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

8 0
3 years ago
Read 2 more answers
Stronger acids have more _______
mario62 [17]
Examples of strong acids are hydrochloric acid (HCl), perchloric acid (HClO4), nitric acid (HNO3) and sulfuric acid (H2SO4). ... For example, hydrogen chloride is a strong acid in aqueous solution, but is a weak acid when dissolved in glacial acetic acid.
7 0
3 years ago
The internal pressure on a bottle of soda is 5550 torr. the bottle cap is designed to withstand a pressure of 7.2 atm. (a) calcu
Mademuasel [1]

This is a conversion problem. 760 Torr is equal to 1 atmosphere. so we can can solve now. 5,550torr*(1atmosphere/760torr)=7.3atmospheres. since the two torrs cancel out with cross multiplication we are left in atmosphere units.

6 0
3 years ago
Read 2 more answers
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