## 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)=?
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
Physics-based and algebra-based solution
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
as x = v₀ t + ½at²
then x = √p t + ¼qt²

⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾ⁱⁿ₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎ₐₑₒₓₔₕₖₗₘₙₚₛₜ·ΔΘΣΦΧΨΩαβγδεζηθικλμνξπρςστφχψωϑϓϕ√∛∜∞∠∣∥∫∬∭∮∯≤≥⊥℃ℇ℉ℏ℔№⅀⅏⎛⎞⎜⍊⎝½

## Saturday, October 8, 2016

### Blackboard, Physical Problem

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## 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 P in terms of variables: the mass M, the final speed V, and the distance d.
P
=P(M,V,d) ?

Solution:
Pₐ=W/t
K-Kₒ=W
Kₒ=0
KMV²
PₐMV²/t
V=at
d=½at²= ½Vt
t=2d/V

PₐMV²/(2d/V) = ¼MV³/d

## Friday, September 30, 2016

### Physics Problem with Solution - Acceleration Varies with Time

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## Tuesday, September 20, 2016

### Sunrize

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