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3 years ago

What kind of system does not allow matter or energy to enter or exit?

Physics

1 answer:

Vilka [71]3 years ago

6 0

a closed system does not allow matter or energy to pass through

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A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second

Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

$E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 = \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC$

Thus, the value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

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2 years ago

The bending of waves as they pass between media is called .?

Rufina [12.5K]

Refraction is the change in direction of waves that occurs when waves travel from one medium to another. Refraction is always accompanied by a wavelength and speed change. Diffraction is the bending of waves around obstacles and openings.

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2 years ago

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Diabetes occurs when there is a large amount of sugar in the blood. Which gland is most likely not functioning well enough in a

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C pancreas! i think it should be

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2 years ago

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is: