Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S =
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Answer : The final velocity of the ball is, 12.03 m/s
Explanation :
By the 3rd equation of motion,
where,
s = distance covered by the object = 6.93 m
u = initial velocity = 2.99 m/s
v = final velocity = ?
a = acceleration =
Now put all the given values in the above equation, we get the final velocity of the ball.
Thus, the final velocity of the ball is, 12.03 m/s
Answer:
16 minutes
Explanation:
This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.
The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."
Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...
(4 ft)/(0.25 ft/min) = 16 min
The level in the two tanks will be the same after 16 minutes.
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<em>Additional comment</em>
The oil levels at that time will be 4 ft.
You can write two equations for height:
y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)
y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)
These will be equal when ...
y = y
12 -0.5x = 8 -0.25x
4 = 0.25x . . . . . . . . . . add 0.5x -8
16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height
The graph shows when the tanks will have equal heights and when they will be drained.
Answer:
a) 159.07 MPa
b) 10.45 MPa
c) 79.535 MPa
Explanation:
Given data :
length of cantilever beam = 1.5m
outer width and height = 100 mm
wall thickness = 8mm
uniform load carried by beam along entire length= 6.5 kN/m
concentrated force at free end = 4kN
first we determine these values :
Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m
Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N
A) determine max bending stress
б = =
= 159.07 MPa
B) Determine max transverse shear stress
attached below
ζ = 10.45 MPa
C) Determine max shear stress in the beam
This occurs at the top of the beam or at the centroidal axis
hence max stress in the beam = 159.07 / 2 = 79.535 MPa
attached below is the remaining solution
