The ocean may pull away rom the shore as a tsunami approaches?
2 answers:
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Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v =
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀
Answer:
Explanation:
Given that
d= 1.5 in ( 1 in = 0.0254 m)
d= 0.0381 m
P= 75 hp ( 1 hp = 745.7 W)
P= 55927.5 W
N= 1800 rpm
We know that power P is given as
T=Torque
N=Speed
T=296.85 N.m
The maximum shear stress is given as
We know that 1 MPa =0.145 ksi
Answer:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
