Answer:
Δ L = 2.57 x 10⁻⁵ m
Explanation:
given,
cross sectional area = 1.6 m²
Mass of column = 26600 Kg
Elastic modulus, E = 5 x 10¹⁰ N/m²
height = 7.9 m
Weight of the column = 26600 x 9.8
= 260680 N
we know,
Young's modulus=
stress =
=
= 162925
strain =
now,
Δ L = 2.57 x 10⁻⁵ m
The column is shortened by Δ L = 2.57 x 10⁻⁵ m
Answer: F(t) = 11 - 0.9(t)
Explanation:
We know the following:
The candle burns at a ratio given by:
Burning Ratio (Br) = 0.9 inches / hour
The candle is 11 inches long.
To be able to create a function that give us how much on the candle remains after turning it after a time (t). We will need to know how much of the candle have been burned after t.
Let look the following equation:
Br = Candle Inches (D) / Time for the Candle to burn (T) (1)
Where (1) is similar to the Velocity equation:
Velocity (V) = Distance (D)/Time(T)
This because is only a relation between a magnitude and time.
Let search for D on (1)
D = Br*T (2)
Where D is how much candle has been burn in a specif time
To create a function that will tell us how longer remains of the candle after be given a variable time (t) we use the total lenght minus (2):
How much candle remains? ( F(t) ) = 11 inches - Br*t
F(t) = 11 - 0.9(t)
F(t) defines the remaining length of the candle t hours after being lit
Answer:
The periodic table illustrate some of the elements from Hydrogen to Calcium
B, because neutrons contain no charge, therefore dont affect the charge of the atom
Answer:
a) T2 = 133.5°C
x2 = 0.364
b) Qout = 3959.6 kJ
Explanation:
For this exercise we will make two assumptions: the tank is stationary, the kinetic and potential energy are equal to zero. The second assumption is that there are interactions between the two tanks. The contents of both tanks will be a closed system. The energy balance equals:
ΔEsystem = Ein - Eout
-Qout = ΔUA + ΔUB = (m*(u2-u1))A + (m*(u2-u1))B
The steam properties for the two tanks in the initial state can be found in Table A-4 to table A-6 of Cengel:
P1A = 1000 kPa
T1A = 300°C
v1A = 0.25799 m^3/kg
u1A = 2793.7 kJ/kg
T1B = 150°C
x1 = 0.5
vf = 0.001091 m^3/kg
uf = 631.66 kJ/kg
vg = 0.39248 m^3/kg
ufg = 1927.4 kJ/kg
v1B = vf + x1*vfg = 0.001091 + (0.5*(0.39248-0.001091)) = 0.19679 m^3/kg
u1B = uf + x1*ufg = 631.66 + (0.5*1927.4) = 1595.4 kJ/kg
V = VA + VB = mA*v1A + mB*v1B = (2*0.25799) + (3*0.19679) = 1.106 m^3
m = mA + mB = 3+2 = 5 kg
the specific volume will be equal to:
v2 = V/m = 1.106/5 = 0.2213 m^3/kg
With these calculations, we can looking the new properties in the same tables:
P2 = 300 kPa
v2 = 0.2213 m^3/kg
T2 = Tsat, 300 kPa = 133.5°C
x2 =(v2-vf)/(vg-vf) = (0.22127-0.001073)/(0.60582-0.001073) = 0.364
u2 =uf + x2*ufg = 561.11 + (0.364*1982.1) = 1282.8 kJ/kg
-Qout = (2*(1282.8-2793.7)) + (3*(1282.8-1595.4)) = -3959.6 kJ
Qout = 3959.6 kJ