A horizontal wire is stretched with a tension of 98.4 N, and the speed of transverse waves for the wire is 406 m/s. What must th
1 answer:
Answer:
Amplitude = 4.725 mm
Explanation:
We have got the following values :
T = 98.4 N
T is the wire tension
v = 406 m/s
v is the transverse waves speed
f = 60.5 Hz
f is the frequency
ΔU = 0.391 W
ΔU is the average power carried by the wave
If μ is the mass per unit length of the wire ⇒ μ =
If ω is the angular frequency of the wire ⇒ ω = 2π.f
So ω = 2π(60.5 Hz) and ω units are
The equation that relates all is :
ΔU = (1/2).(T/v^2).(ω^2).(A^2).v
Where A is the wave amplitude
This will give us A in meters
You might be interested in
Answer:
Explanation:
This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".
In the y-stuff category:
v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)
Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)
a = -9.8 m/s/s
t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)
In the "x-stuff" category:
v₀ = 7.25 m/s (this is given)
Δx = ???
a = 0 (acceleration in this dimension is ALWAYS 0)
t = (we will solve for this in the y-dimension and plug it in here).
In the y dimension:
Δx = v₀t + and plugging in from the y-dimension info:
which simplifies to
so
which, to 2 significant digits is
t = 1.4 seconds
Now we will do the same in the x-dimension, using t = 1.4:
Δx = v₀t + and filling in the x-stuff:
Δx = Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:
Δx = 7.25(1.4) + 0 so
Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).
Answer:
1.44 x 10⁻⁶ C
Explanation:
= charge on one sphere
= charge on other sphere
= Total charge on the two spheres = 40 μC
=
= 40 x 10⁻⁶
= (40 x 10⁻⁶) -
eq-1
= distance between the two spheres = 50 cm = 0.50 m
= magnitude of force between the two spheres = 2.0 N
Magnitude of force between the two spheres is given as
= 1.44 x 10⁻⁶ C