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Maslowich
2 years ago
5

Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy

stem. have the Sun at their exact center. are highly inclined to the ecliptic. are almost circular, with low eccentricities.
Physics
1 answer:
BaLLatris [955]2 years ago
4 0

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

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Can a body possess velocity at the same time in horizontal and vertical directions?​
iogann1982 [59]

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

5 0
1 year ago
Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster
Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2

Dividing by g

\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g

The acceleration is 1.10261 times g

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s

In mph

186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph

The speed of the dragster is 416.17506 mph

5 0
2 years ago
Urban cities like Atlanta have to contend with a serious problem like pollution. Drivers in California are testing out a car tha
riadik2000 [5.3K]

1. With this statement, the author is referring to the fact that the vehicles are one of the largest polluters of the air. In order to reduce the pollution, the vehicles that are used will need to be changed, and with it the pollution will decrease significantly. The reduction of the pollution will come because the vehicles on hydrogen will not cause any pollution, so the enormous amounts of carbon dioxide released from the combustion of the engines will be thing of the past.

2. There are several challenges with this type of vehicles in order for them to replace the fossil fuel driven ones. The big price is one of the factors, as the majority of the people can not afford these cars. Another problem is that these vehicles are not as fast as the fossil fuel driven ones, and lot of people enjoy fast driving, despite it not being safe. There are millions of vehicles out there on the roads, and changing all of them with hydrogen vehicles will take a lot of time, as lot of those vehicles are new ones, so the people will not be willing to just throw them away and leave them rot in their garages. In order for the change of the driving park to be accomplished, the prices should go down, the people to be more serious about the environment and its protection, and patience as several decades will probably be needed for a change like this to be competed.

3. The fuel celled cars are a viable answer to decrease the pollution, as they are not causing any pollution, but instead will stop the process of large emissions of carbon dioxide from the fossil fueled cars. While this method is a good one, it should not be the only, as on its own it can not have the desirable effect, but instead all the major polluters should be included in the process. The industry and the production of energy are one of the major polluters as well, so they will need to follow the example, as if they not, the problem will stay, considering that the industry is constantly growing and the demand for energy is constantly growing too.

4 0
2 years ago
A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8  + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

    T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

 T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0
2 years ago
for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha
VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

R=\dfrac{u^2\ \sin2\theta}{g}

Here,

u is the speed of launch of projectile

Here, \theta=30^{\circ}

We need to find the other launch angle when the projectile have the same range, such that,

\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}

\sin(60)=\sin2\alpha

\dfrac{\sqrt3}{2}=\sin2\alpha

\alpha =30^{\circ}

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0
2 years ago
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