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damaskus [11]
3 years ago
6

PLEASE HELP ME

Physics
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

the time period is 22.63 years

Explanation:

The computation of the number of years taken is as follows:

According to the Keplers law

T^2 \propto R^3

Here T denotes the time period

And, R denotes the average distance among the sun and planet

Now let us assume for earth

R_1 & T_1 be 1 year

So, for Planet

R_2 = 8R_1 , T_2 = ?

Now applied Keplers law

(\frac{T_2}{T_1})^2 \propto (\frac{R_2}{R_1})^3\\\\(T_2)^2 = (\frac{8R_1}{R_1})^3 \times 1^2\\\\(T_2)^2 = 512\\\\T_2 = 22.674

So, the time period is 22.63 years

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In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220-m-long cannon with final speed of 10.
Sidana [21]

Answer:

The unrealistically large acceleration experienced by the space travelers during their launch is 2.7 x 10⁵ m/s².

How many times stronger than gravity is this force? 2.79 x 10⁴ g.

Explanation:

given information:

s = 220 m

final speed, vf = 10.97 km/s = 10970 m/s

g = 9.8 m/s²

he unrealistically large acceleration experienced by the space travelers during their launch

vf² = v₀²+2as, v₀ = 0

vf² = 2as

a =vf²/2s

  = (10970)²/(2x220)

  = 2.7 x 10⁵ m/s²

Compare your answer with the free-fall acceleration

a/g = 2.7 x 10⁵/9.8

a/g = 2.79 x 10⁴

a = 2.79 x 10⁴ g

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3 years ago
What is electric current
Bond [772]

Answer:

Electric current is electric charge in motion. It can take the form of a sudden discharge of static electricity, such as a lightning bolt or a spark between your finger and a ground light switch plate. ... Most electric charge is carried by the electrons and protons within an atom.

Explanation:

because it is

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3 years ago
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

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  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
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       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
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vesna_86 [32]

Answer:

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Two particles are separated by a certain distance. the force of gravitational interaction between them is f0. now the separation
12345 [234]
 <span>as i recall, gravity is relative to the square of the distance. 

so if the distance is tripled, then the gravitational attraction would be reduced by 3^2 or 1/9. 
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</span>
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3 years ago
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