Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
Answer:
Answer for the question:
Use the Hurricanes data and via Multiple regression select the three input variables: Min-pressure, Gender, Category in order to predict the All-Deaths. Pick one of the three variables that has the best P-value and re-do the regression using ONLY this one variable to predict the All-deaths. Answer the questions. (Numerical answers are rounded so choose the answer that matches the best).
is explained in the attachment.
Explanation:
Answer:
Temperature distribution is
Heat flux=q
Heat rate=q A
Explanation:
We know that for no heat generation and at steady state
a and are the constant.
Given that heat flux=q
We know that heat flux given as
From above we can say that
Alos given that when x= L temperature is T(L)=T
So temperature T(x)
So temperature distribution is
Heat flux=q
Heat rate=q A (where A is the cross sectional area of wall)
Answer:
420.69 mg
Explanation:
i just did that last week
Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.
Answer:
Height of the water = √(128)/147456 ft
Explanation:
Given
Radius, r = 1 ft
Height, h = 2 ft
Radius of hole = 1/32in
Acceleration of gravity, g = 32ft/s²
c = 0.6
Area of the hold = πr²
A = π(1/32)² ---- Convert to feet
A = π(1/32 * 1/12)²
A = π/147456 ft²
Area of water = πr²
A = π 1²
A = π
The differential equation is;
dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water
A1 = π/147456, A2 = π
dh/dt = (π/147456)/π √(2*32*2)
dh/dt = 1/147456 * √128
dh/dt = √128/147456 ft
Height of the water = √(128)/147456 ft