Answer:
The acceleration of the satellite is 
Explanation:
The acceleration in a circular motion is defined as:
(1)
Where a is the centripetal acceleration, v the velocity and r is the radius.
The equation of the orbital velocity is defined as
(2)
Where r is the radius and T is the period
For this particular case, the radius will be the sum of the high of the satellite (
) and the Earth radius (
) :
Then, equation 2 can be used:
⇒ 
Finally equation 1 can be used:
Hence, the acceleration of the satellite is
: You could make slight changes to the process or consider wheather the experiment was carried out correctly
Close the switch would be the correct answer
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
