Answer:
T = 4.42 10⁴ N
Explanation:
this is a problem of standing waves, let's start with the open tube, to calculate the wavelength
λ = 4L / n n = 1, 3, 5, ... (2n-1)
How the third resonance is excited
m = 3
L = 192 cm = 1.92 m
λ = 4 1.92 / 3
λ = 2.56 m
As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio
v = λ f
f = v / λ
f = 343 / 2.56
f = 133.98 Hz
Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m
The expression for standing waves on a string is
λ = 2L / n
λ = 2 0.37 / 2
λ = 0.37 m
The speed of the wave is
v = λ f
As we have some resonance processes between the string and the tube the frequency is the same
v = 0.37 133.98
v = 49.57 m / s
Let's use the relationship of the speed of the wave with the properties of the string
v = √ T /μ
T = v² μ
T = 49.57² 18
T = 4.42 10⁴ N
Force on a moving charge is given by formula
here we know that this force will be maximum when velocity is perpendicular to magnetic field
here we know that
now we have
The initial force of the throw overcomes gravity quite easily. Then, gravity begins to bring it back down to earth, making a curved path.
Answer: 1.77 s
Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:
xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then
(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s
then by solving the quadratric equation in t
t=1.77 s
P1v1/t1 = p2v2/t2
p1=p2, v1=.2, t1=333, t2=533
we can find v2 from this
be aware, temperature must be in Kelvin.
