Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h
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Answer
given,
y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]
length of the rope = 1.33 m
mass of the rope = 3.31 g
comparing the given equation from the general wave equation
y(x,t)= A cos[k x+ω t]
A is amplitude
now on comparing
a) Amplitude = 2.20 mm
b) frequency =
f = 118.25 Hz
c) wavelength
d) speed
v = 105.84 m/s
e) direction of the motion will be in negative x-direction
f) tension
T = 27.87 N
g) Power transmitted by the wave
P = 0.438 W
Answer:
z = 3,737 10⁵ m
Explanation:
a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation
P V = n R T
P = (n r / V) T
We replace
P = (n R / V) T₀
b) Let's apply this equation in the points
Lower
.z = 0
P₀ = 610 Pa
P₀ = (nR / V) T₀
Higher.
P = 10 Pa
P = (n R / V) T₀ e^{- C z}
We replace
P = P₀ e^{- C z}
e^{- C z} = P / P₀
C z = ln P₀ / P
z = 1 / C ln P₀ / P
Let's calculate
z = 1 / 1.1 10⁻⁵ ln (610/10)
z = 3,737 10⁵ m