1.) Pitch
2.)Wavelength
3.)Density/Elastic Properties-b. Two of the above
4.)Liquids
5.) I'm not sure but I'm pretty sure it's the Doppler effect
6.) Frequency Increases
It is true because <span>A pyramid of biomass is a representation of the amount of energy contained in biomass, at different trophic levels for a given point in time . The amount of energy available to one trophic level is limited by the amount stored by the level below. Because energy is lost in the transfer from one level to the next, there is successively less total energy as you move up trophic levels. Tree is a base as it provides food and energy.</span>
There are 3 significant figures, if that answers the question.
Answer:
1.195 m
2.8375 s
2.21433 rad/s
Explanation:
d = Distance = 2.39 m
N = Number of cycles = 8
t = Time to complete 8 cycles = 22.7 s
Radius would be equal to the distance divided by 2

The radius is 1.195 m
Time period would be given by

Time period of the motion is 2.8375 s
Angular speed is given by

The angular speed of the motion is 2.21433 rad/s
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