Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.
Answer:
See the attached picture for answer.
Explanation:
See the attached picture for explanation.
Answer:
b). Occurs at the outer surface of the shaft
Explanation:
We know from shear stress and torque relationship, we know that

where, T = torque
J = polar moment of inertia of shaft
τ = torsional shear stress
r = raduis of the shaft
Therefore from the above relation we see that

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.
When r= 0, then τ = 0
and when r = R , τ is maximum
Thus, torsional shear stress is maximum at the outer surface of the shaft.
Answer:
43248 newtons.
Explanation:
Force = mass x accelerations and units of force are newtons which are given in the question.
here mass = 125 of air and 2.2 of fuel, total = 125+2.2=127.5kg/s and the velocity of the exhaust is 340m/s.
force = 340m/s * 127.5kg/s = 43248 newtons technically this is wrong (observe units) but i will expalin how i have taken acceleration as a velocity here and mass/unit time as simply mass.
see force is mass times acceleration or deceleration, here our velocity is not changing therefore it is constant 340m/s but if it were to change and become 0 in one second then there would be -340m/s^2 (note the units ) of deceleration and there would be force associated with it and that force is what i have calculated here. similarly there would be mass in flow rate of mass per second, which is also in that one second of time.
let's calculate error.
error = (actual-calculated)/actual. = (43248-60000)/43248= -38.734% less is ofcourse greater than 2%.
So the load cell is not reading correct to within 2% and it should read 43248newtons.
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft