Answer:
no of unit is 17941
Explanation:
given data
fixed cost = $338,000
variable cost = $143 per unit
fixed cost = $1,244,000
variable cost = $92.50 per unit
solution
we consider here no of unit is = n
so here total cost of labor will be sum of fix and variable cost i.e
total cost of labor = $33800 + $143 n ..........1
and
total cost of capital intensive = $1,244,000 + $92.5 n ..........2
so here in both we prefer cost of capital if cost of capital intensive less than cost of labor
$1,244,000 + $92.5 n < $33800 + $143 n
solve we get
n > 
n > 17941
and
cost of producing less than selling cost so here
$1,244,000 + $92.5 n < 197 n
solve it we get
n >
n > 11904
so in both we get greatest no is 17941
so no of unit is 17941
The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec
<h3>How to find the displacement?</h3>
We are given the velocity equation as;
s' = 40 - 3t²
Thus, the speed equation will be gotten by integration of the velocity equation to get;
s = ∫40 - 3t²
s = 40t - ¹/₂t³
Thus, the displacement between times of t = 2 sec and t = 4 sec is;
∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]
∆S = 210 m
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Given Information:
Initial temperature of aluminum block = 26.5°C
Heat flux = 4000 w/m²
Time = 2112 seconds
Time = 30 minutes = 30*60 = 1800 seconds
Required Information:
Rise in surface temperature = ?
Answer:
Rise in surface temperature = 8.6 °C after 2112 seconds
Rise in surface temperature = 8 °C after 30 minutes
Explanation:
The surface temperature of the aluminum block is given by

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.
After t = 2112 sec:

The rise in the surface temperature is
Rise = 35.1 - 26.5 = 8.6 °C
Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.
After t = 30 mins:

The rise in the surface temperature is
Rise = 34.5 - 26.5 = 8 °C
Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.
Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>
Answer:
The amplitude of the absorbed mass can be found
for ka:

now
![w^2=\frac{K_{a} }{m_{a} } \\m_{a} =\frac{K_{a} }{w^2} =\frac{125000}{[6000*2\pi /60]^2} =0.317kg](https://tex.z-dn.net/?f=w%5E2%3D%5Cfrac%7BK_%7Ba%7D%20%7D%7Bm_%7Ba%7D%20%7D%20%5C%5Cm_%7Ba%7D%20%3D%5Cfrac%7BK_%7Ba%7D%20%7D%7Bw%5E2%7D%20%3D%5Cfrac%7B125000%7D%7B%5B6000%2A2%5Cpi%20%2F60%5D%5E2%7D%20%3D0.317kg)