Answer:
meter per second
Explanation:
It could be any other unit such as yard or feet, put it will be whatever measure per second or whatever time.
Examples
feet per second
miles per hour
I'll be happy to solve the problem using the information that
you gave in the question, but I have to tell you that this wave
is not infrared light.
If it was a wave of infrared, then its speed would be close
to 300,000,000 m/s, not 6 m/s, and its wavelength would be
less than 0.001 meter, not 12 meters.
For the wave you described . . .
Frequency = (speed) / (wavelength)
= (6 m/s) / (12 m)
= 0.5 / sec
= 0.5 Hz .
(If it were an infrared wave, then its frequency would be
greater than 300,000,000,000 Hz.)
Answer:
the correct answer is C
Explanation:
When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.
Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.
Applying this to our case if there is
10 boulder on the chart
in reality there are #_boulder = 10 30
#_boulder = 300 boulder
so the correct answer is C
Answer
given,
I is the loudness of sound
I = 10 Log₁₀ r
r is relative intensity
at when relative intensity is 10⁶
I = 60 dB
how much louder when 100 people would be talking together
I = 10 Log₁₀ r
I = 10 Log₁₀ (10⁶ x 100)
I = 10 Log₁₀ (10⁸)
I = 80 dB
hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.
Use the eq. of Young modulus Y=(F/A)/(∆l/lo)
dimana ∆l is the elongation of wire, lo is its initial length.
So ∆l = (F/A)lo/Y.
∆l = (1000N/(6.5 × 10^-7 m^2))×(2.5m)/(2.0 × 10^-11 N/m^2)
Use calculator to finish it.
