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emmasim [6.3K]
2 years ago
11

One of the foci for the moon's orbit would be the

Physics
2 answers:
Likurg_2 [28]2 years ago
6 0

Answer:

2) Earth

Explanation:

As per Kepler's first law we know that all planets and all satellites revolves in elliptical path with one of its focii as Sun or planet around which satellite and planets are revolving.

Now here we need to know about the focii of moon which is revolving around Earth in its elliptical path.

So as per same law we can say that since moon is revolving around the Earth so its focii must be Earth

So here correct answer would be

2) Earth

suter [353]2 years ago
5 0
The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated.  You know that motion is
always relative to something, and the solar system is not simple.
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A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
2 years ago
Which bright solar feature is shown in the picture above?
Ghella [55]

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called <em>Prominence</em>.

<em>Photosphere</em> is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

6 0
3 years ago
Read 2 more answers
Se colocan tres objetos, muy cerca uno del otro, dos al mismo tiempo. Cuando se juntan los objetos A y B, se repelen. Cuando se
docker41 [41]

Answer:

Los objetos A y C tienen cargas del mismo signo (opcion a)

Explanation:

Hay dos tipos de cargas : cargas positivas y cargas negativas.

La Ley de Coulomb dice que la fuerza electrostática entre dos cargas puntuales es proporcional al producto de las cargas e inversamente proporcional al cuadrado de la distancia que las separa, y tiene la dirección de la línea que las une y se cumple que:

  • La fuerza ejercida sobre una carga apunta hacia la otra cuando las dos tienen distinto signo (fuerza atractiva).
  • El sentido de la fuerza se dirige hacia el lado opuesto de la carga cuando ambas tienen el mismo signo (fuerza repulsiva).

Es decir que las cargas de igual signo se repelen, mientras que las de diferente signo se atraen.

Entonces, si se juntan los objetos A y B y se repelen significa que la carga es del mismo signo.

Cuando se acercan los objetos B y C, se repelen. Entonces significa que posee carga de igual signo.

Por lo que podes concluir que <u><em>los objetos A y C tienen cargas del mismo signo (opcion a)</em></u>

8 0
2 years ago
"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3
avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

We know that magnetic field due to long wire is given by B=\frac{\mu _0i}{2\pi r}

It is given that magnetic field is zero at some distance so

\frac{\mu _0i_1}{2\pi x}=\frac{\mu _0i_2}{2\pi (24-x)}

Here i_1=3\ A \ and\  i_2=8\ A

So \frac{3}{x}=\frac{8}{24-x}=6.5454\ m

3 0
3 years ago
An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

4 0
3 years ago
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