Answer:
0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +
Explanation:
Given the wave equation of the form :
y(x, t) = ym sin(kx ± ωt)
Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m
Tension, T = 10 N
Amplitude, A = 0.12 mm
Frequency, F = 100 Hz
Comparing with the general wave equation :
y = Asin(kx ± ωt)
A = amplitude = ym = 0.12 mm
2.) k = 2π / λ
Recall :
v = fλ
v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472
λ = v/ f = 4.472 / 100 = 0.04472
Hence,
k = (2 * π) / 0.04472
k = 140.50 rad/m
3.) Angular frequency, ω
ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec
4.) sign is +ve
Direction of wave propagation as given is in the negative x axis
For 2, i think that the reason the ultrasound can make an image is by using the waves that it can use to pick up (???)
idk, that's what i think anyway, i hope i helped ^^
A change of colour, heat being released or fizzing (a gas being released)
The image is missing, so i have attached it.
Answer:
The force exerted on the large block by the small block = 8.4 N
Explanation:
From the image attached, the mass of the small block = 2M while the mass of the large block = 3M
Also,Force on small block = F and force on large block = 2F
Equilibrium of forces on the left gives;
2F - N = 3Ma
Thus,
Ma = (2F - N)/3 - - - - eq1
Also, on right hand side, Equilibrium of forces gives;
N - F = 2Ma
Ma = (N - F)/2 - - - - eq2
Equating eq(1) and eq(2) gives us;
(2F - N)/3 = (N - F)/2
Where N is the force exerted on the large block by the small block.
Making N the subject gives;
4F - 2N = 3N - 3F
5N = 7F
N = 7F/5
We are given F = 6N
Thus;
N = 7(6)/5
N = 8.4 N