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3 years ago

Suppose the sphere is electrically neutral. Is it attracted to

Physics

1 answer:

Elden [556K]3 years ago

8 0

Answer:

No, it is not attracted.

Explanation:

If any sphere is electrically neutral it is not attracted. The materials which are attracted by magnet are called magnetic material whereas which are not attracted are called non magnetic material. Sphere made up of non magnetic materials such as glass, wood, paper will not attracted weather is kept near north pole or near south pole.

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What is the net force in this image?

9966 [12]

Net force on the image is 0 Newton.

<h3>What is force?</h3>

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Net upward force is balanced by net downward force so net force is zero.

Net force on the image is 0 Newton.

To learn more about force refer to the link:

brainly.com/question/13191643

#SPJ1

6 0

1 year ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

What are some of the ways of reducing your background sources of radiation

galben [10]

Radiation is a form of energy traveling through space (air) as particles or waves ..

I hope this help you

7 0

3 years ago

¿Qué proporción deberían guardar los platos de una prensa hidráulica para que, aplicando 40N de fuerza en el plato menor, podamo

9966 [12]

Answer:

El área de la placa más grande es aproximadamente 9,81 veces el tamaño del área de la placa más pequeña

Explanation:

Los parámetros dados son;

La fuerza aplicada sobre la placa más pequeña, F = 40 N

El culo del objeto levantado en la placa más grande, m = 40 kg

El peso del objeto, W = 40 kg × 9,81 m / s² = 392,4 N

Tenemos la presión en el plato pequeño = La presión en el plato más grande

Por lo tanto;

F / A₁ = W / A₂

Dónde;

A₁ = El área del plato pequeño

A₂ = El área de la placa más grande

Por lo tanto, obtenemos;

40 N / (A₁) = 392,4 N / (A₂)

A₂ / A₁ = 392,4 N / (40 N) = 9,81

∴ A₂ = A₁ × 9,81

El área de la placa más grande, A₂ = 9,81 (9,81 m / s² ≈ g) multiplicado por el área de la placa más pequeña A₁.

3 0

3 years ago

What is the maximum centripetal acceleration experienced by a person standing still on the surface of the Earth? Where must they

skelet666 [1.2K]

Answer:

The person must be located in the Equator Line. The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

Explanation:

Physically speaking, the centripetal acceleration ( $a_{r}$ ), measured in meters per square second, experienced by a person is defined by the following expression:

$a_{r} = \omega^{2}\cdot r$ (1)

Where:

$\omega$ - Angular speed of the Earth, measured in radians per second.

$r$ - Distance perpendicular to the rotation axis, measured in meters.

Since rotation axis passes through poles and distance described above is directly proportional to centripetal acceleration. The person must be located in the Equator Line, which is equivalent to the radius of the planet.

In addition, the angular speed of the Earth can be calculated in terms of its period ( $T$ ), measured in seconds:

$\omega = \frac{2\pi}{T}$ (2)

If we know that $r = 6.371\times 10^{6}\,m$ and $T = 86400\,s$ , then the maximum centripetal acceleration experienced by a person is:

$a_{r} = \left(\frac{2\pi}{86400\,s} \right)^{2}\cdot (6.371\times 10^{6}\,m)$

$a_{r} = 0.0337\,\frac{m}{s^{2}}$

The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

6 0

3 years ago