If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of e
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Answer:
The correct answer is D /
= 1.5
Explanation:
In this exercise the moment of inertia equation should be used
I = ∫ r² dm
In addition to the parallel axis theorem
I = + M D²
Where is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest
Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in
L = 100.00 cm (1m / 100 cm) = 1.0000 m
= 50 cm = 0.50 m
Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density
λ = M / L = dm / dx
dm = λ dx
Let's replace in the moment of inertia equation
I = ∫ x² ( λ dx)
We integrate
I = λ x³ / 3
We evaluate between the lower limits x = -L/2 to the upper limit x = L/2
I = λ/3 [(L/2)³ - (-L/2)³] = lam/3 [L³/8 - (-L³ / 8)]
I = λ/3 L³/4
I = 1/12 λ L³
Let's replace the linear density with its value
I = 1/12 (M/L) L³
I = 1/12 M L²
Let's calculate with the given values
I = 1/12 M 1²
I = 1/12 M
This point is the center of mass of the rod
Icm = I = 1/12 M = 8.333 10-2 M
Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is
D = - x
D = 0.50 - 0.30
D = 0.20 m
Let's calculate
=
+ M D²
= 1/12 M + M 0.202
= M (1/12 + 0.04)
= M 0.123
To find the relationship between the two moments of inertia, divide the quantities
/
= M 0.123 / (M 8.3 10-2)
/
= 1.48
The correct answer is d 1.5
Answer:
3349J/kgC
Explanation:
Questions like these are properly handled having this fact in mind;
- Heat loss = Heat gained
Quantity of heat = mcΔ∅
m = mass of subatance
c = specific heat capacity
Δ∅ = change in temperature
m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)
m₁ = mass of block = 500g = 0.5kg
c₁ = specific heat capacity of unknown substance
∅₂ = block initial temperature = 50oC
∅₁ = equilibrium temperature of block and water after mix= 25oC
m₂= mass of water = 2kg
c₂ = specific heat capacity of water = 4186J/kg C
∅₃ = intial temperature of water = 20oC
0.5c₁(50-25) = 2 x 4186(25-20)
And we can find c₁ which is the unknown specific heat capacity
c₁ = = 3348.8J/kg C≅ 3349J/kg C