Answer:
i)ω=3600 rad/s
ii)V=7059.44 m/s
iii)F=1245.8 N
Explanation:
i)
We know that angular speed given as

We know that for one revolution
θ=2π
Given that time t= 2 hr
So
ω=θ/t
ω=2π/2 = π rad/hr
ω=3600 rad/s
ii)
Average speed V

Where M is the mass of earth.
R is the distance
G is the constant.
Now by putting the values


V=7059.44 m/s
iii)
We know that centripetal fore given as

Here given that m= 200 kg
R= 8000 km
so now by putting the values


F=1245.8 N
The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays
<h3>How to utilize z-score statistics?</h3>
We are given;
Mean; μ = 15
Standard Deviation; σ = 5
We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.
Formula for z-score is;
z = (x' - μ)/σ
We want to find the value of x such that the probability is 0.95;
P(X > x) = P(z > (x - 15)/5) = 0.95
⇒ 1 - P(z ≤ (x - 15)/5) = 0.95
Thus;
P(z ≤ (x - 15)/5) = 1 - 0.95
P(z ≤ (x - 15)/5) = 0.05
The value of z from the z-table of 0.05 is -1.645
Thus;
(x - 15)/5 = -1.645
x ≈ 7
Complete Question is;
A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.
Read more about Z-score at; brainly.com/question/25638875
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Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In (
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61