Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :
f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
The correct answer for the question that is being presented above is this one: "<span>c. Planets orbit in elliptical patterns; a planet's orbit covers equal areas in equal amounts of time; planets' orbits are shorter or longer depending on their distance from the Sun."</span>
Here are the following choices:
a. Planets orbit in elliptical patterns; the bigger the planet, the more gravitational pull; a planet's gravitational pull is stronger or weaker depending on its distance from the Sun.
b. A planet's orbit covers equal distances in equal amounts of time; the speed of a planet's orbit depends on its distance from the Sun; the bigger the planet, the slower it moves.
c. Planets orbit in elliptical patterns; a planet's orbit covers equal areas in equal amounts of time; planets' orbits are shorter or longer depending on their distance from the Sun.
The answer is
6.67 Ohms (I tried to find the ohms sign but it looks like the lululemon logo)
Mike enters a revolving door that is not moving. Mike should
push at the edge of the door where it is largest distance from the pivot point
in order to produce a torque with the least amount of force. Torque is equal to
t = force x distance.
The answer I had got was C x
