Answer:
The answer is D.
Explanation:
As shown in the picture, the measuring tube is collecting 10 cm of rain.
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, = 39 kHz
velocity of the bat, 
= 8.32 m/s
speed of sound in air, v = 340 m/s
The apparent frequency of sound striking the wall is calculated as;
The frequency reflected by the stationary wall to the bat is calculated as;
Part 1
When the solar atmosphere accumulates a lot of magnetic energy
to a point that cannot accumulate more, all that magnetic energy is suddenly released,
and with it, a lot of radiation. So much, that in fact it covers all of the
electromagnetic spectrum; from radio waves to gamma rays. That burst of
radiation is called a solar flare. In a single solar flare the amount of
radiation released is millions of times greater than all the nuclear bombs in
the face if the earth exploding together. Lucky for us, most of the high-energy
radiation dissipates before reaching the Earth, and the radiation that do reach
us, is deflected by the Earth’s magnetic field.
Part 2
1. Not all the radiation
of solar flares that reach the Earth is deflected by its magnetic field; some
of them reach us and charges the upper atmosphere with ionized particles. Those
particles react with the gases in the atmosphere and produce a light; that
light is what we call Auroras borealis or southern nights; One the most beautiful
natural spectacles in earth, who thought Auroras begin their lives as deadly
solar flares.
2. Solar flares
contain a lot of high-energy radiation that is extremely dangerous for our
electronic devices; when they reach the Earth, they can damage sensible
electronics like satellites. A very powerful solar flare could even damage all
the electronic devices on the surface of the Earth.
<span>P = energy/t = 0.0025/1E-8 = 250000 W
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2
I(peak) = 2I(ave) = 8.8113463E11 W/m^2
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
From the figure, you can observe that a/b < 1, thus E < W
