Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
<em>weight = (mass) x (gravity)</em>
Weight = (5.00 kg) x (9.81 m/s²)
weight = (5.00 x 9.81) (kg-m/s²)
<em>Weight = 49.05 Newton</em>
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
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