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Natalka [10]
3 years ago
8

Which of the following statement describes an actual orbit

Physics
2 answers:
Leno4ka [110]3 years ago
7 0

I suppose right answer is d because staellite means an object that move around the larger object and Jupiter also moves around the Sun

Papessa [141]3 years ago
6 0

Answer:

The Moon is a satellite of Earth

Explanation:

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1. A record with a radius of 0.3m spins in a clockwise circle with a centripetal
Hitman42 [59]

Solve for the linear/tangential speed:

<em>a</em> = <em>v</em>²/<em>r</em>

where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.

4.7 m/s² = <em>v</em>²/(0.3 m)

<em>v</em>² = (0.3 m) (4.7 m/s²)

<em>v</em> ≈ 3.96 m/s

For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is

(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s

Take the reciprocal of this to get the period:

1 / (7.46 rev/s) ≈ 0.134 s/rev

So it takes the record about 0.134 seconds to complete one revolution.

6 0
3 years ago
A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?
tester [92]

Answer:

Explanation:

Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.

I = 0.5 mr²

Total kinetic energy, T = 0.5 mv² + 0.5 Iω²

T = 0.5 (mv² + 0.5 mr²ω²)

v = rω

So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²

Rotational kinetic energy is

R = 0.5 Iω² = 0.5 x 0.5 mr²ω²

R = 0.25 mv²

So, R / T = 0.25 / 0.75 = 1/3

5 0
3 years ago
An acorn falls from a tree. Its velocity just before it hits the ground is 28.2 m/s, downward. (acceleration of gravity is 9.81m
ss7ja [257]

Answer:

12.74 ms^-1 download

Explanation:

v=28.2, a=9.81

start from rest u=0

v=u+at=0+(9.81)t=28.2

t=2.875...

it reach 1.4 second before hitting the ground:

t=1.4, u=0, a=9.81

v=u+at=0+(9.81)(1.4)=12.74

7 0
3 years ago
spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center o
satela [25.4K]

Answer:

the correct result is r = 3.71 10⁸ m

Explanation:

For this exercise we will use the law of universal gravitation

          F = - \frac{m_{1} m_{2} }{r^2}

We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m

rocket force -Earth

          F₁ = - \frac{m' M }{r^2}

rocket force - Moon

          F₂ = - \frac{m' m }{(d-r)^2}

in the problem ask for what point the force has the relation

          2 F₁ = F₂

let's substitute

          2 2 \frac{M}{r^2} = \frac{m}{(d-r)^2}

          (d-r) ² = \frac{m}{2M} r²

           d² - 2rd + r² = \frac{m}{2M} r²

           r² (1 -\frac{m}{2M}) - 2rd + d² = 0

Let's solve this quadratic equation to find the distance r, let's call

           a = 1 - \frac{m}{2M}

           a = 1 - \frac{7.36 10^{22} }{2 \  5398 10^{24}} = 1 - 6.15 10⁻³

           a = 0.99385

         

            a r² - 2d r + d² = 0

           r =  \frac  {2d \frac{+}{-}   \sqrt{4d^2 - 4 a d^2}} {2a}

           r = [2d ± 2d \sqrt{1-a}] / 2a

           r = \frac{d}{a}   (1 ± √ (1.65 10⁻³)) =  \frac{d}{a} (1 ± 0.04)

           r₁ = \frac{d}{a} 1.04

           r₂ = \frac{d}{a} 0.96

let's calculate

           r₁ = \frac{3.84 10^8}{0.99385} 1.04

           r₁ = 401.8 10⁸ m

          r₂ = \frac{3.84 10^8}{0.99385} 0.96

          r₂ = 3.71 10⁸ m

therefore the correct result is r = 3.71 10⁸ m

3 0
2 years ago
A positive charge is placed in an electric field that points west. What direction is the force on the positive particle? I think
Kisachek [45]
Right.  You are true.  The direction of the electric field is defined to be
the direction of the force on a small positive charge placed in the field.
3 0
3 years ago
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