IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron
The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.
A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.
It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:
-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .
Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency 
Velocity of the string wave is equal to 
Power of wave propagation is equal to 
So power of the wave will be equal to 5.464 watt
Explanation:
well there is nothing there and it could be different by diffrent objects, idk
(a) Let's convert the final speed of the car in m/s:

The kinetic energy of the car at t=19 s is

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:

But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into

(c) The instantaneous power is given by

where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:

And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s: