A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
P is 525 lb, determine the values of the angle a and the required magnitude of the vertical force, F.
1 answer:
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
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Answer:
Explanation:
We need only to apply the definition of acceleration, which is:
In our case the final velocity is , the initial velocity is
since it departs from rest, the final time is
and the initial time we are considering is
So for our values we have:
Answer:
So the specific heat of the liquid B is greater than that of A.
Explanation:
Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,
where:
m = mass of the body
c = specific heat of the body
change in temperature of the body
The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.
So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.
So the specific heat of the liquid B is greater than that of A.