Answer:
0.2687 approximately 0.27
Explanation:
Diameter = 0.320
Speed = 40.0 rev/min
We are required to find coefficient of static friction between friction and button
The radius can be calculated as
0.320/2
= 0.160m
Then we have the rotational speed w = 40rev/min x 2pi/60
= 4.19 rad/s
umg = mrw²
u = mrw²/mg
u = rw²/g -------(1)
g = 9.8
When we put values into equation 1
0.150m x 4.19² / 9.8
= 0.150m x 17.5561 /9.8
= 0.2689
This is approximately 0.27
Answer:
Temperature of water leaving the radiator = 160°F
Explanation:
Heat released = (ṁcΔT)
Heat released = 20000 btu/hr = 5861.42 W
ṁ = mass flowrate = density × volumetric flow rate
Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³
ṁ = 1000 × 0.000126 = 0.126 kg/s
c = specific heat capacity for water = 4200 J/kg.K
H = ṁcΔT = 5861.42
ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C
And in change in temperature terms,
10°C= 18°F
11.08°C = 11.08 × 18/10 = 20°F
ΔT = T₁ - T₂
20 = 180 - T₂
T₂ = 160°F
Answer:
Explanation:
Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.
Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t
A(0) = 10 g
Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min
The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min
But the concentration is total amount of salt over 350L constant volume
C = A / 350
Therefore our rate of change for salt A' is
A' = 5 - 5A/350 = 5 - A/70
This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is
So
with A(0) = 10
c + 350 = 10
c = 10 - 350 = -340
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