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Snezhnost [94]
3 years ago
5

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 work

is required to bring an additional electron from infinity to the origin?
Physics
1 answer:
aksik [14]3 years ago
5 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{q_e q_e}{d^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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A particle moving along the y-axis has the potential energy u =4y3j, where y is in m. what is the y-component of the force on th
Ivenika [448]

Potential energy is given as

U = 4y^3

now as we know that force is related by potential energy by the formula

F = - \frac{dU}{dy}

So it is gradient of energy with position in Y

F = - \fracd(4y^3}{dy}

F = -12y^2 \hat j

Now at y = 0

F = 0 N

at y = 1

F = - 12*1^2

F = - 12 N

at y = 2

F = - 12*2^2

F = - 48 N

so above is the forces at given positions

7 0
3 years ago
Which symbol represents a type of radiation that has the same mass as an
Korolek [52]

Answer:β^+

Explanation: it is beta + , also positron emission radiation

5 0
2 years ago
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A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the waterme
Nonamiya [84]

Answer:

Work, W = 846.72 Joules

Explanation:

Given that,

Mass of the watermelon, m = 4.8 kg

It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.

W = mgh

W=4.8\ kg\times 9.8\ m/s^2\times 18\ m

W = 846.72 Joules

So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.

7 0
3 years ago
A sled is slowing down at the bottom of a snowy hill.
Kitty [74]

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

5 0
2 years ago
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

  • ∑F is net force
  • m is mass of the third child
  • a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

  • the wagon
  • the children outside the wagon

<h3>Free body diagram</h3>

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

<h3>Acceleration of the  child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

7 0
2 years ago
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