Answer: Cumulus
Explanation: Most large cloud fronts are made up of cumulus clouds, large storm clouds are cumulonimbus clouds.
In one of the most amazing coincidences in all of science,
the part of the electromagnetic spectrum that's visible to the
human eye is called "visible light".
Visible light is not 'divided' into anything. We mention the names
to seven of the colors in visible light. But all of the thousands of
OTHER colors that we can see are in there too, even though we
don't bother to list their names when we buzz through the rainbow
in the third grade.
Answer:
this the way that hydraulic pump works:
The system of hydraulic jack lifts the heavy objects by transferring the oil from one chamber to another.
Through the suction valves into the two cylinders which creates pressure distributing the fluid equally to the points causing the objects to lift up.
hope these points helped u
any confusion then comment it
I think your question should be:
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is
What is the rms value of (a) the electric field and
(b) the magnetic field in the electromagnetic wave emitted by the laser
Answer:
a) 
b) 
Explanation:
To find the RMS value of the electric field, let's use the formula:
Where
;
;
Therefore
![E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}](https://tex.z-dn.net/?f=%20E_r_m_s%20%3D%20sqrt%2A%7B%281.239%2A10%5E9W%2Fm%5E2%29%20%2F%20%5B%283.00%2A10%5E8m%2Fs%29%2A%288.85%2A10%5E-%5E1%5E2C%5E2%2FN.m%5E2%29%5D%7D%20)
b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:
;
;
Answer:
B. 1700 Hz, 5100 Hz
Explanation:
Parameters given:
Length of ear canal = 5.2cm = 0.052 m
Speed of sound in warm air = 350 m/s
The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:
f(m) = m * (v/4L)
Where m is an odd integer;
v = velocity
L = length of the tube
Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).
f(1) = 1 * [350/(4*0.052)]
f(1) = 1682.69 Hz
Approximately, f(1) = 1700 Hz
f(3) = 3 * [350/(4*0.052)]
f(3) = 5048 Hz
Approximately, f(3) = 5100 Hz
