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3 years ago

How are energy and distance up the ramp related

1 answer:

8 0

I don't exactly understand the question but I'll try my best to answer it and help you. You need energy in order to go up the ramp and cover the distance. I hope this helps! :)

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A point charge of magnitude q is at the center of a cube with sides of length L.a) What is the electric flux through each of the

agasfer [191]

<h2>The flux through each face is q/6ε₀ .</h2>

Explanation:

The charge q is placed at the center of the cube of side L

According to Gauss's law the flux through any closed surface is q/ε₀

here q is the charge enclosed .

In this case cube has the six faces . The flux through each face = q/6ε₀

In the second case The cube has the face with length L₁

The flux through each face = q/6ε₀

Thus flux through the cube does not depend upon the size of the cube .

3 0

3 years ago

Describe how you would make or construct a wet cell. What basic part would you need? How would you use them to produce a voltage

polet [3.4K]

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4 0

3 years ago

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitationa

sdas [7]

Answer:

The correct answer is option B)

Explanation:

Considering the given question as -

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.

We know that gravitational pull (F) between any two bodies of mass $M_{1}$ and $M_{2}$ is given by -

F = $\dfrac{GM_{1}M_{2} }{r^{2} }$ where 'r' is the distance between the two bodies.

Let ,

$M_{e}$ : Mass of the earth

$M_{m}$ : Mass of the moon

m : Mass of the satellite

$r_{e}$ : Distance of satellite from earth

$r_{m}$ : Distance of satellite from moon

Given that $r_{e}$ = $r_{m}$

Let $r_{e}$ = $r_{m}$ =r

Force on satellite by the earth is -

$F_{e}$ = $\dfrac{GM_{e}m }{r^{2} }$

Force on satellite by the moon is -

$F_{m}$ = $\dfrac{GM_{m}m }{r^{2} }$

∵ Mass of earth ( $M_{e}$ ) > Mass of moon ( $M_{m}$ )

∴ $F_{e}$ > $F_{m}$

∴ The gravitational pull of earth on satellite is more than that of the moon.

4 0

4 years ago

La cantidad de materia en un objeto.

Scrat [10]

La respuesta correcta es Masa

Explicación:

Todos los objetos existentes en el universo se componen de materia cuya unidad básica es conocida como átomo. A su vez el átomo se compone de subpartículas que incluyen protones, neutrones y electrones. Además de esto, la cantidad de materia o átomos en un objeto o cuerpo, por ejemplo la cantidad de materia en un balón se conoce como masa. Este factor se mide en unidades como gramos o kilos, por ejemplo, la cantidad de masa en una manzana es de aproximadamente de 150 gr. Finalmente, la masa se diferencia del peso porque en el peso la fuerza gravitacional y no solamente la materia se debe considerar.

7 0

3 years ago

The Earth is 81.25 times as massive as the Moon and the radius of the Earth is 3.668 times the radius of the Moon. If a simple p

Alexeev081 [22]

Answer:

option (b)

Explanation:

Mass of moon = m

Mass of earth = 81.25 x mass of moon = 81.25 m

Radius of moon = r

radius of earth = 3.668 x radius of moon = 3.668 r

Frequency on earth = 2 Hz

Let the frequency on moon is f.

The formula for the frequency is given by

$f = \frac{1}{2\pi }\times \sqrt{\frac{g}{l}}$

The value of acceleration due to gravity on earth is g.

ge = G Me / Re^2

ge = G x 81.25 m / (3.668 r)^2

ge = 6.039 x G m / r^2 = 6.039 x gm

ge / gm = 6.039

Now use the formula for frequency

$\frac{fe}{fm} = \sqrt{\frac{ge}{gm}}$

$\frac{2}{fm} = \sqrt{\frac{6.039 gm}{gm}}$

$\frac{2}{fm} = 2.46$

fm = 0.814 Hz

4 0

3 years ago