Answer:
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Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Answer: condensation.
Vaporization is the pass from liquid state to gaseous state.
Then the reverse is the transformation from gaseous state to liquid state.
That is called condensation.
When the water vaporizes the liquid transforms into vapor which goes to the atmosphere. When the water vapor of the atmosphere condensates liquid water is formed. You can see condensation when you have a glass with cold water and drops of water form in the exterior of the glass: those drops are liquid water that formed when the vapor of the air that surrounds the glass cools due to the lower temperature of the surface of the glass.
Answer:
Explanation:
In a quiet forest, you can sometimes hear a single leaf fall to the ground. ... greater its pressure amplitude, the more the air is compressed in the sound it creates. ... Graphs of the gauge pressures in two sound waves of different intensities. ... The sound intensity level β in decibels of a sound having an intensity I in watts per .
Answer:
0.173 m.
Explanation:
The fundamental frequency of a closed pipe is given as
fc = v/4l .................. Equation 1
Where fc = fundamental frequency of a closed pipe, v = speed of sound l = length of the pipe.
Making l the subject of the equation,
l = v/4fc ................ Equation 2
also
v = 331.5×0.6T ................. Equation 3
Where T = temperature in °C, T = 18.0 °c
Substitute into equation 3
v = 331.5+0.6(18)
v = 331.5+10.8
v = 342.3 m/s.
Also given: fc = 494 Hz,
Substitute into equation 2
l = 342.3/(4×494)
l = 342.3/1976
l =0.173 m.
Hence the length of the organ pipe = 0.173 m.
