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2 years ago

Name TWO WEAKNESSES of the model pictured below

Physics

1 answer:

Nikolay [14]2 years ago

7 0

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare, and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

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Please answer ASAP, and please don't joke around and actually do answer my question. I will give you brainliest if you answer it

hammer [34]

Answer:

30 feet

Explanation:

First off, if they are throwing at 12.0 m/s and it takes 2.5 seconds. It will be the act of multiplication.

12 times 2.5 is 30, because 12 times 2 is 24 plus 12 divided by 2 which is 6 so 24 plus 6 is 30.

8 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

A rat runs 2m right, turns around and runs 3m left. Then goes 2m right. What is its displacement?

geniusboy [140]

Answer:

Explanation:

I think this answer would be 1m to the left.

7 0

2 years ago

What must be true the number of chromosomes in each sex cell

zvonat [6]

Answer:

In humans, each cell normally contains 23 pairs of chromosomes, for a total of 46.

8 0

3 years ago

if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a

Whitepunk [10]

Answer:

$r=0.41m$

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

$\tau=r\times F$

Due to the definition of cross product, the magnitude of the torque is given by:

$\tau=rFsin\theta$

Where $\theta$ is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when $sin\theta$ is equal to one, solving for r:

$r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m$

7 0

2 years ago