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slava [35]
3 years ago
8

Now put the names of the planets in increasing order based on their distance from the sun

Physics
1 answer:
Anika [276]3 years ago
5 0
The eight planets of the Solar System arranged in order from the sun:

Mercury: 46 million km / 29 million miles (.307 AU)
Venus: 107 million km / 66 million miles (.718 AU)
Earth: 147 million km / 91 million miles (.98 AU)
Mars: 205 million km / 127 million miles (1.38 AU)
Jupiter: 741 million km /460 million miles (4.95 AU)
Saturn: 1.35 billion km / 839 million miles (9.05 AU)
Uranus: 2.75 billion km / 1.71 billion miles (18.4 AU)
Neptune: 4.45 billion km / 2.77 billion miles (29.8 AU)

Astronomers often use a term called astronomical unit (AU) to represent the distance from the Earth to the Sun.

+ Pluto (Dwarf Planet): 4.44 billion km / 2.76 billion miles (29.7 AU)
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If an astronaut had a mass of 30 kg on the moon what would his mass be on earth?
Novay_Z [31]

Explanation:

A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon

hope it help..... pls add me as brainlist.

Have a nice day

7 0
3 years ago
The movement of a magnetic pole away from the actual pole
Lunna [17]

The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole.
3 0
3 years ago
A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
erastova [34]
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps




5 0
3 years ago
An incident on route 12 summary?I need to describe what happens in the book in a short summary can anyone help?
Serga [27]

Answer:  An Incident on Route 12 is presented here in a high quality paperback edition. This popular classic work by James H. Schmitz is in the English language, and may not include graphics or images from the original edition.

Explanation: I HOPE THAT HELPED

5 0
3 years ago
A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re
arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car =  distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a* t^2)/2

distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0
3 years ago
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