A pendulum is not a wave.
-- A pendulum doesn't have a 'wavelength'.
-- There's no way to define how many of its "waves" pass a point
every second.
-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.
-- The frequency of a pendulum depends only on the length
of the string from which it hangs.
If you take the given information and try to apply wave motion to it:
Wave speed = (wavelength) x (frequency)
Frequency = (speed) / (wavelength) ,
you would end up with
Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz
Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ? That's pretty absurd.
This math is not applicable to the pendulum.
Answer:
No.
Explanation:
A feather is less dense and thus less force exerted while a rock is very dense thus exerting more force .
Answer:
P_(pump) = 98,000 Pa
Explanation:
We are given;
h2 = 30m
h1 = 20m
Density; ρ = 1000 kg/m³
First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,
Thus, it can be expressed as;
P_(tank)+ P_(pump) = P_(nozzle)
Now, the pressure would be given by;
P = ρgh
So,
ρgh_1 + P_(pump) = ρgh_2
Thus,
P_(pump) = ρg(h_2 - h_1)
Plugging in the relevant values to obtain;
P_(pump) = 1000•9.8(30 - 20)
P_(pump) = 98,000 Pa
Volumetric flasks are most accurate
Answer: the photograph will likely show only one star.
Explanation:
Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.
