Answer:
The answer ix below
Explanation:
There are 12 apple pies left.
Given that:
n = number of apple pies left = 12
x = number of food banks = 6
1) For the 12 apple pies to be distributed among 6 food banks. The number of ways in which this can be done is:
C(n + x - 1, x - 1) = C(12 + 6 -1, 6 - 1) = C(17, 5) = [tex]\frac{17!}{(17-5)!5!} =\frac{17!}{12!5!}=6188 \ ways[/tex]
12 apple pies can be distributed among 6 food banks in 6188 ways
2) For the 12 apple pies to be distributed among 6 food banks if each food bank must receive one pie, 6 pies would be remaining. The number of ways in which this can be done is:
C((n - x) + x - 1, n - x) = C(12 - 6 + 6 - 1, 12 - 6) = C(11, 6) = [tex]\frac{11!}{(11-6)!6!} =\frac{11!}{6!5!}=462 \ ways[/tex]
12 apple pies can be distributed among 6 food banks if each food bank must receive one pie in 462 ways
A single-threaded 25-mm power screw hasa pitch of 5 mm. The frictional diameter of the collar is 45 mm. The max load onvertical direction of the screw is5kN. The collar has a coefficients of friction of0.06, and he threads hasa coefficients of friction of0.09. Find the overall efficiency and the torque to "raise" and "lower" the load.
Answer:
torque to raise the load = 16.411 Nm
torque to lower the load = 8.40 Nm
overall efficiency = 0.24
Explanation:
Given:
max load on vertical direction of the screw = Force = F = 5kN
frictional diameter of the collar = 45 mm
Diameter = 25 mm
length of pitch = 5 mm
coefficient of friction for thread µ = 0.09
coefficient of friction for collar µ[tex]_{c}[/tex] = 0.06
To find:
torque to "raise" the load
torque to and "lower"
overall efficiency
Solution:
Compute torque to raise the load:
[tex]T_{R} = \frac{ Fd_{m}}{2} (\frac{L+(\pi ud_{m}) }{\pi d_{m}-uL }) +\frac{Fu_{c} d_{c} }{2}[/tex]
where
[tex]T_{R}[/tex] is the torque
F is the load
[tex]d_{m}[/tex] is diameter of thread
[tex]d_{c}[/tex] is diameter of collar
L is the thread pitch distance
µ is coefficient of friction for thread
µ[tex]_{c}[/tex] is coefficient of friction for collar
Putting the values in above formula:
[tex]T_{R}[/tex] = 5(25) / 2 [5+ (π(0.09)(25) / π(25)-0.09(5)] + 5(0.06)(45) / 2
= 125/2 [5 + (3.14)(0.09)(25)/ 3.14(25)-0.45] + 13.5/2
= 62.5 [(5 + 7.065) / 78.5 - 0.45] + 6.75
= 62.5 [12.065 / 78.05 ] + 6.75
= 62.5 (0.15458) + 6.75
= 9.66125 + 6.75
= 16.41125
[tex]T_{R}[/tex] = 16.411 Nm
Compute torque to lower the load:
[tex]T_{L} = \frac{ Fd_{m}}{2} (\frac{(\pi ud_{m}) - L }{\pi d_{m}-uL }) +\frac{Fu_{c} d_{c} }{2}[/tex]
= 5(25) / 2 [ (π(0.09)(25) - L / π(25)-0.09(5) ] + 5(0.06)(45) / 2
= 125/2 [ ((3.14)(0.09)(25) - 5) / 3.14(25)-0.45 ] + 13.5/2
= 62.5 [ (7.065 - 5) / 78.5 - 0.45 ] + 6.75
= 62.5 [ 2.065 / 78.05 ] + 6.75
= 62.5 (0.026457) + 6.75
= 1.6535625 + 6.75
= 8.40 Nm
Since the torque required to lower the the load is positive indicating that an effort is applied to lower the load, Hence the thread is self locking.
Compute overall efficiency:
overall efficiency = F(L) / 2π [tex]T_{R}[/tex]
= 5(5) / 2(3.14)( 16.411)
= 25/ 103.06108
overall efficiency = 0.24
Some STEM occupations use (blank) who assist and support the lead personnel with projects or experiments.
Answer:
technicians
Explanation:
Just answer this question with this answer...
I got it right so you should too...
Answer:
Technicians
Explanation:
What friction rate should be used to size a duct for a static pressure drop of 0.1 in wc if the duct has a total equivalent length of 150 ft
Answer:
0.067wc
Explanation:
The formula is actual static pressure loss = (total equivalent divided by 100) multiplied by rate of friction
We substitute values
actual static pressure = 0.1
Total equivalent length = 150 ft
0.1 = (150ft/100) multiplied by Rate of friction
Friction rate at 100ft = 0.067
So we have that the required friction needed is 0.067wc
An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output
Answer:
The rate of work output = -396.17 kJ/s
Explanation:
Here we have the given parameters
Initial temperature, T₁ = 355°C = 628.15 K
Initial pressure, P₁ = 350 kPa
h₁ = 763.088 kJ/kg
s₁ = 4.287 kJ/(kg·K)
Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately
h₂ = 79.572 kJ/kg
The saturation temperature at the given
T₂ = 79°C
The rate of work output [tex]\dot W[/tex] = [tex]\dot m[/tex]×[tex]c_p[/tex]×(T₂ - T₁)
Where;
[tex]c_p[/tex] = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)
[tex]\dot m[/tex] = The mass flow rate = 2.0 kg/s
Substituting the values, we have;
[tex]\dot W[/tex] = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s
[tex]\dot W[/tex] = -396.17 kJ/s
Water of dynamic viscosity 1.12E-3 N*s/m2 flows in a pipe of 30 mm diameter. Calculate the largest flowrate for which laminar flow can be expected (in lpm). What is the corresponding flowrate if it is an air flow (1.8E-5 N*s/m2 )
Answer:
For water
Flow rate= 0.79128*10^-3 Ns
For Air
Flow rate =1.2717*10^-3 Ns
Explanation:
For the flow rate of water in pipe.
Area of the pipe= πd²/4
Diameter = 30/1000
Diameter= 0.03 m
Area= 3.14*(0.03)²/4
Area= 7.065*10^-4
Flow rate = 7.065*10^-4*1.12E-3
Flow rate= 0.79128*10^-3 Ns
For the flow rate of air in pipe.
Flow rate = 7.065*10^-4*1.8E-5
Flow rate =1.2717*10^-3 Ns
what is the total inductance of a circuit that contains two 10 uh inductors connected in a parallel?
Answer:
5 microhenries
Explanation:
The effective value of inductors in parallel "add" in the same way that resistors in parallel do. The value is the reciprocal of the sum of the reciprocals of the inductances that are in parallel.
10 uH ║ 10 uH = 5 uH
The effective inductance is 5 uH.
Give four effects of water hammer.
Explanation:
The hammer effect (or water hammer) can harm valves, pipes, and gauges in any water, oil, or gas application. It occurs when the liquid pressure is turned from an on position to an off position abruptly. When water or a liquid is flowing at full capacity there is a normal, even sound of the flow.
Complete the grading of fine aggregate table given below. Plot grading curve and calculate
fineness modulus. Also comment on the type of the grading curve.
Answer:
Attachment...?
Explanation: