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.
<span>the gravational potential energy of anything on the ground is zero. When calculating potential energy you take height in meters and multiply it by the mass of the object in kilograms and the acceleration of gravity to get a new unit called Joules.
Any object at ground level has a potential energy of zero newtons becuase anything multiplied by zero is zero. An object with mass of 54 kg, 4 meters above the ground has a gravitatinal potential energy of 2116.8 Joules.</span>
Explanation:
Acceleration is defined as the change in velocity over time.
When there is an increment or increase in the magnitude of velocity of a moving body then it is known as positive acceleration.
Whereas when there is a decrease in magnitude of velocity of a moving body then it is known as negative acceleration.
Thus, we can conclude that positive acceleration occurs when an object speeds up.
Answer: A
<u>Explanation:</u>
NOTES:
d = 650 meters
t = 10 seconds
**********************************
v = d/t
= 650 meters/10 seconds
= 65 meters/second
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
