Wednesday, December 21, 2016
Sunday, November 27, 2016
Find the Water Speed V₂ in the Pipe on the Top Floor
Water at a gauge pressure P₁ at street level (zero, reference height) flows into an office building through a pipe with the radius R₁ of the circular cross-section. The pipe tapers down to the radius R₂ of the circular cross-section by the top floor, located at the height h, where the faucet has been left open. Find the water speed V₂ in the pipe on the top floor in terms of P₁, R₁, R₂, h, water density ρ and g. Assume no branch pipes and ignore viscosity.
Find the Vertical Support Forces at Ends of the Lower Steel Beam
The
lower uniform steel beam has a mass M.
Only one-third of an identical steel beam is resting on the lower
steel beam, as shown in Fig. Find the vertical support forces at ends
of the lower steel beam in terms of M
and g.
t₀=0, x₀=0, v(x)=√(p+qx). x(t)=?
t₀=0
x₀=0
v(x)=√(p+qx)
x(t)=?
x₀=0
v(x)=√(p+qx)
x(t)=?
Calculus-based solution:
dx/dt=q⁰·⁵(p/q+x)⁰·⁵
(p/q+x)⁻⁰·⁵dx = q⁰·⁵dt
(p/q+x)⁻⁰·⁵d(p/q+x) = q⁰·⁵dt
2(p/q+x)⁰·⁵ = q⁰·⁵t + c
2(p/q+x)⁰·⁵ -2(p/q+x₀)⁰·⁵= q⁰·⁵(t-t₀)
2(p/q+x)⁰·⁵ -2(p/q)⁰·⁵= q⁰·⁵t
(p/q+x)⁰·⁵ -(p/q)⁰·⁵= ½q⁰·⁵t
(p/q+x)⁰·⁵ = ½q⁰·⁵t+(p/q)⁰·⁵
p/q+x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵ + p/q
x = ¼qt² + q⁰·⁵t(p/q)⁰·⁵
x = p⁰·⁵t + ¼qt²
x = v₀ t + ½at²
v₀=p⁰·⁵
a=½q
v₀=p⁰·⁵
a=½q
Physics-based and algebra-based solution
If a=const then v²=v₀²+2a·Δx, so
v=√(v₀²+2a·Δx)
If a=const then v²=v₀²+2a·Δx, so
v=√(v₀²+2a·Δx)
As it is given:
v=√(p+qx)
so p=v₀², v₀=√p=p⁰·⁵
as x₀=0 then 2a·Δx=2ax
2ax=qx, 2a=q, a=½q
v=√(p+qx)
so p=v₀², v₀=√p=p⁰·⁵
as x₀=0 then 2a·Δx=2ax
2ax=qx, 2a=q, a=½q
as x = v₀ t + ½at²
then x = √p t + ¼qt²
then x = √p t + ¼qt²
⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ·ΔΘΣΦΧΨΩαβγδεζηθικλμνξπρςστφχψωϑϓϕ√∛∜∞∠∣∥∫∬∭∮∯≤≥⊥℃ℇ℉ℏ℔№⅀⅏⎛⎞⎜⍊⎝½
Tuesday, October 25, 2016
A Rocket is Moving in Outer Space
A rocket is #moving in #outer #space where no #external #forces exist. #rocket #IsMoving #outerspace #externalforces #exist #movingrocket pic.twitter.com/VOgcB6nYLA
— TutorState.com (@TutorStateCom) October 25, 2016
Friday, October 14, 2016
Saturday, October 8, 2016
Saturday, October 1, 2016
Average Power
What
is the average power output of an engine when a car of mass M
accelerates uniformly (a=const)
from rest to a final speed V
over a distance d
on flat, level ground?
Ignore energy lost due to friction and air resistance.
Derive a formula for average power Pₐ in terms of variables: the mass M, the final speed V, and the distance d.
Pₐ=Pₐ(M,V,d) ?
Ignore energy lost due to friction and air resistance.
Derive a formula for average power Pₐ in terms of variables: the mass M, the final speed V, and the distance d.
Pₐ=Pₐ(M,V,d) ?
Solution:
Pₐ=W/t
K-Kₒ=W
Kₒ=0
K=½MV²
Pₐ=½MV²/t
V=at
d=½at²=
½Vt
t=2d/V
Pₐ=½MV²/(2d/V)
= ¼MV³/d
Friday, September 30, 2016
Tuesday, September 20, 2016
Wednesday, July 13, 2016
Tuesday, June 28, 2016
Wednesday, June 8, 2016
Sunday, May 15, 2016
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