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3 years ago

¿Qué sucede con la Energía interna de un sistema cuando sobre él se efectúa trabajo? ¿Qué ocurre con su temperatura?

Physics

1 answer:

disa [49]3 years ago

5 0

Answer:

Disminuye la energía y la temperatura internas.

Explanation:

translation:

it decreases the internal energy and temperature

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A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction

makkiz [27]

Answer:

0.1 N

Explanation:

using F = ma yields 0.1 N.

net

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3 years ago

In two or more complete sentences, explain how you can prove that the number of degrees that the Moon rotates around the Earth e

sasho [114]

A close representation of the moons motion around earth is that it moves around the earth once in 27.3 days. the moons average movement is 13.2 degrees per day so around 92 per week.

Hope this helps!

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3 years ago

Each of the rods depicted below were machined from same stock metal. They were originally machined to be the same length, but th

Shkiper50 [21]

The force required to extend a rod increases as the cross sectional area

increases.

The rod that experiences the largest force is <u>rod B</u>

Reason:

The elongation of a rod by the application of a force is given by the

following formula;

$\Delta L = \dfrac{F \cdot L}{A \cdot E}$

From the above equation, we have that the elongation is inversely

proportional to the cross sectional area, such that the extension of a rod by

a given force reduces as the cross sectional area of the rod increases.

Therefore, the force required to extend the length of a rod by a specific

amount increases as the cross sectional area of the rod increases,

indicating that the rod with the largest cross sectional area require the

most force and therefore, experiences the largest force.

The rod that experiences the largest force is the rod with the largest cross

sectional area, which is <u>rod B</u>

Learn more here:

brainly.com/question/12937199

4 0

2 years ago

Hola! necesito las respuestas de "fuerzas"

Alexus [3.1K]

I need help on this to I have to get the answer fast

5 0

3 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago