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2 years ago

As two nuclei are moved closer

Physics

1 answer:

Alexus [3.1K]2 years ago

4 0

Answer: increases

Explanation:

As two nuclei are moved closer together, the electrostatic force of

repulsion between them increases

According to the law of electrostatic which state that unlike charges attract while like charges repel each other.

The proton in the two nuclei are positively charged which tend to repel each other.

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Draw the vector C⃗ =1.5A⃗ −3B⃗ . The length and orientation of the vector will be graded. The location of the vector is not impo

Nutka1998 [239]

I made the drawing in the attached file.

I included two figures.

The upper figure shows the effect of:

- multiplying vector A times 1.5.
It is drawn in red with dotted line.

- multiplying vector B times - 3 .
It is drawn in purple with dotted line.

In the lower figure you have the resultant vector: C = 1.5A - 3B.

The method is that you translate the tail of the vector -3B unitl the point of the vector 1,5A, preserving the angles.

Then you draw the arrow that joins the tail of 1,5A with the point of -3B after translation.

The resultant arrow is the vector C and it is drawn in black dotted line.

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The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

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Rx= 3.5 km

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) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force

Nana76 [90]

Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others.
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
$F_x=F_3+F_2cos(45^{\circ})$
$F_y=F_1+F_2sin(45^{\circ})$
Let's figure out each component:
$F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\$
$F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}$
Total force acting on the charge would be:
$F=\sqrt{F_x^2+F_y^2}$
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
$F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N$
$F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N$
Now we can find the total force acting on a single charge:
$F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N$
As said before, intensity of the force acting on charges is the same for all of them.

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2 years ago

The ball of dough hits the floor and does not rebound.

tekilochka [14]

Answer:

When the ball goes down its mechanical energy is conserved, ust before touching the ground all the energy is kinetic

When the ball touches the floor, energy has been converted into potential and heat, by the deformation of the ball.

Explanation:

When the ball goes down its mechanical energy is conserved, this is the power energy due to the height it is converted into kinetic energy to medicad that falls, just before touching the ground all the energy is kinetic.

When the ball touches the floor, the kinetic energy is not conserved, but if we define a system formed by the ball and the floor, the amount of movement is conserved, this being an inelastic shock, because the bla and the floor are stuck, so which energy has been converted into potential and energized and heat by the deformation of the ball.

Consequently all the mechanical energy that the ball brings before reaching the ground was converted into potential energy and heat during the crash.

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2 years ago