What happens when a proton is brought near another proton?

What's the thrust force of 20,000N

nika2105 [10]

Answer:

895g

Explanation:

5 0

3 years ago

A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh

maksim [4K]

Answer: $16.234^{\circ}C$

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water( $m_w$ )=600 gm

temperature of water =15

specific heat of water $=4.186 j/g-^{\circ}C$

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.) $=mgh=6\times 9.81\times 10=588.6 J$

Heat supplied $=m_wc(\Delta T)$

$H.E.=600\times 4.186\times (T-16)$

$588.6=2511.6\times (T-16)$

T-16=0.234

$T=16.234^{\circ}C$

This is not an efficient way of heating water as there is only $0.234^{\circ}C$ increase in temperature.

5 0

3 years ago

A car's gas tank contains 58.7 kg

Bogdan [553]

Answer:

721 kg/m^3

Explanation:

Trust me bro

8 0

3 years ago

Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar

den301095 [7]

Answer:

electric field Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

a) Let's write the electric field for each charge and the total field

E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

Et = E1 + E2 + E3

E1 = k q / (x-a)²

E2 = k (-2q) / x²

E3 = k q / (x + a)²

Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ where x << 1, for this we take factor like x from all the equations

Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

(1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

(1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0

2 years ago

Johnson made a hole at the bottom of a plastic bottle containing water. However, he noticed that the water did not flow out from

Sphinxa [80]

Answer:

a) See explanation below

b) At X.

Explanation:

Please, see the picture attached with the image of the plastic bottle for this question.

<u>(a) Explain why the water could not flow out of the bottle.</u>

What makes the water flow out of the botlle is the force of gravity, whic attracts the water towards the Earth.

When Johnson made a small hole at the bottom of the plastic bottle containing water, the air outside the bottle, which surrounds it and exerts a pressure all over the outer walls of the bottle, exerted a force against the small area of water "over" the hole that is in contact with the air.

Thus, this force of the air pushing upward through the wall opposed the force of gravity pulling downward making the net force zero and the water cannot fall.

<u>(b) To make the water flow out more easily, his teacher suggested making another hole. At which position - X, Y or Z, should he make the 2nd hole in order for the water to flow out the fastest?</u>

You must open the hole at a place where there is not water but air, such that the outer air can enter in the bottle.

That will make that the pressure in the space over the water inside the bottle be equal to the pressure outside.

The pressure of the air above the water will push it downward. Now, the force from the pressure of air inside the water, which is downward, opposes the upward force from the pressure of air around the first hole, and the net force will be downward, making the water flow out more easily.

Thus, the position where he should make the second hole in order for the water to flow fastest is at X.

4 0

3 years ago