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slava [35]
3 years ago
13

Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around th

e Sun. Consider an asteroid that is spherically shaped with radius r and density 2400 kg/m3 .
Part A) You find yourself on the surface of this asteroid and throw a baseball at a speed of 30 m/s . If the baseball is to travel around the asteroid in a circular orbit, what is the largest radius asteroid on which you are capable of accomplishing this feat?

Part B) After you throw the baseball, you turn around and face the opposite direction and catch the baseball. How much time T elapses between your throw and your catch?
Physics
1 answer:
vazorg [7]3 years ago
7 0

Answer:

Lo siento si no entiendes español, pero si lo sabes, necesito estos puntos. Te han robado un punto

Explanation:

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________ occurs when the horizontal winds move air against mountain ranges, forcing air to rise as it passes over the mountains.
Semenov [28]
The answer would be:
Precipitation sometimes occurs when the horizontal winds move air against mountain ranges, forcing air to rise as it passes over the mountains.
This happens when the air is forced to move from low elevation to high elevation due to rising terrain. This causes the air to cool adiabatically. This increases the relative humidity and causes clouds to form, under certain conditions it can also create precipitation.
6 0
3 years ago
X = v xo t. x = (10.0 m/s)(3.53 s) x = ????
LenaWriter [7]

x = V<em>x</em> * t

given V<em>x</em> = 10m/s n t = 3.53s

x = 10 * 3.53

= 35.3m


7 0
3 years ago
Read 2 more answers
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
The higher the amplitude of the wave, the greater its intensity and the greater its loudness.
lyudmila [28]
True,when you turn the volume up on your  television , you're actually turning up the amplitude<span>!

</span>
4 0
3 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
Delvig [45]

Answer:

E=\dfrac{\lambda }{2\pi \varepsilon _or}

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}

E_w=\dfrac{\lambda }{2\pi \varepsilon _or}

Now the electric filed due to hollow metal cylinder

E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}

E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}

Now  by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

E=\dfrac{\lambda }{2\pi \varepsilon _or}

5 0
3 years ago
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