<span>(a).the heat trasfer surface area and heat flux on the surface of filament are
Area of Surface= µDL=3.14(0.05cm)(5cm)= 0.785 cm square
qs=Q/Area of surface= 150W/0.785= 191W/cmsq.=1.91x10Âłx10ÂłW/Msq
(b). the heat surface on the surface of heat bulb
Area of surface = 3.14xD²= 3.14(8CM)²= 201.1cm²
qs=Q/Area of surface=150w/201.1cm²=0.75 w/cm²= 7500w/m²
the amount and cost of electrical energy consumed during one period is
Electrical Consumption=QΛt=(0.15 KW)(365X8h/yr)=438 k Wh/yr
Annual cost= 438 kWh/yr)($.08/kWh)= $ 35.04 /yr</span>
Answer:
a) r eq = -a/(2b)
b) k = a/r eq = -2b
Explanation:
since
U(r) = ar + br²
a) the equilibrium position dU/dr = 0
U(r) = a + 2br = 0 → r eq= -a/2b
b) the Taylor expansion around the equilibrium position is
U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!
,where Un(a) is the nth derivative of U respect with r , evaluated in a
Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative
U(r) = U(r eq) + dU/dr(r eq) (r- r eq) + d²U/dr²(r eq) (r- r eq)² /2
since dU/dr(r eq)=0
U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2
comparing with an energy balance of a spring around its equilibrium position
U(r) - U(r eq) = 1/2 k (r-r eq)² → U(r) = U(r eq) + 1/2 k (r-r eq)²
therefore we can conclude
k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq
thus
k= a/r eq
Answer:
An object moves at a constant speed of 6m/s. this means that the object moves 6 meters every second
Explanation:
Answer:
The phenomenon of total internal reflection of light is used in many optical instruments like telescopes, microscopes, binoculars, spectroscopes, periscopes
The work done on the spring is equal to the elastic potential energy of a spring. We look at Hooke's law, which states that the force needed to stretch a spring is proportional<span> to the displacement of the spring.
</span><span>F = -kx
W = (integral) Fdx
</span>W = (integral) -kx dx
W = kΔx²/2
20 = k(42 - 30) / 2
k = 10/3