The value 100 MW is equivalent to (a) 100×10^6 w (b) 100 x 10^-6 w (c) 100 x 10^-3 w (d) 100 x 10^3 w
1 answer:
Answer:
(a) 100×10⁶ W
Explanation:
Power is defined as the rate of doing work. Thus,
<u>The S.I. unit of Power is watt which is J/s.</u>
Mega is a unit prefix which denotes a factor of 1 million (10⁶). It is denoted by 'M'.
Thus,
<u>1 MW=10⁶ W</u>
The Conversion of 100 MW into W is shown below:
<u>100 MW = 100×10⁶ W</u>
The option that correct resembles is option (a)
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Answer:
Explanation:
Given that :
The initial volume of water = 1000.00 mL = 1000000 mm³
The initial temperature of the water = 10° C
The height of the water column h = 1000.00 mm
The final temperature of the water = 70° C
The coefficient of thermal expansion for the glass is ∝ =
The objective is to determine the the depth of the water column
In order to do that we will need to determine the volume of the water.
We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:
At temperature the density of the water is
At temperature the density of the water is
The mass of the water is
Thus; we can say ;
⇒
Thus, the volume of the water after heating to a required temperature of is 1022400 mm³
However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:
The area of the water before heating is:
The area of the heated water is :
Finally, the depth of the heated hot water is:
Hence the depth of the heated hot water is
Answer:
Test code:
>>u=10;
>>g=9.8;
>>q=100;
>>m0=100;
>>vstar=10;
>>tstar=fzero_rocket_example(u, g, q, m0, vstar)
Explanation:
See attached image
Answer:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Explanation:
Given:
- The diameter of the pipe d = 5-in
- The pipe is supported every L = 9 ft of pipe in length
- The weight if the pipe + contents W = 10 lb/ft
Find:
determine the magnitude and location of the maximum bending moment in members AC and BC.
Solution:
- The figure (missing) is given in the attachment.
- We will first determine the external forces acting on each member:
Section: 9-ft section of pipe.
Sum of forces perpendicular to member AC = 0
F_d - 0.8*W*L = 0
F_d = 0.8*10*9 = 72 lb
Sum of forces perpendicular to member BC = 0
F_e - 0.6*W*L = 0
F_e = 0.6*10*9 = 54 lb
F_d = 72 lb , F_e = 54 lb
- Then we will determine the support reactions for each member AC point A and BC point B.
Section: Entire Frame.
Sum of moments about point B = 0
-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0
-A_y*(1.5625) + 15 + 39.375 = 0
A_y = 34.8 lb
Sum of forces in vertical direction = 0
A_y + B_y - 0.8*F_d - 0.6*F_e = 0
B_y = 0.8*(72) + 0.6*(54) - 34.8
B_y = 55.2 lb
Sum of forces in horizontal direction = 0
A_x + B_x - 0.6*F_d + 0.8*F_e = 0
A_x + B_x = 0
Section: Member AC
Sum of moments about point C = 0
F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0
72*2.5 - 34.8*12 - 9*A_x = 0
A_x = -237.6 / 9 = - 26.4 lb
B_x = - A_x = 26.4 lb
A_x = -26.4 lb , B_x = 26.4 lb
- Now we will calculate bending moment for each member at different sections.
Member AC:
From point A till just before point D
-0.6*A_x*x - A_y*0.8*x + M = 0
15.84*x - 27.84*x + M = 0
M = 12*x ..... max value at D, x = 12.25 in
M_max = 12*12.25/12 = 12.25 lb-ft
Member BC:
From point B till just before point E
-0.8*B_x*x + B_y*0.6*x + M = 0
-21.12*x + 33.12*x + M = 0
M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in
M_max = -12*8.75/12 = -8.75 lb-ft
- The maximum bending moments and their locations are:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
