A drilling operation is performed on a steel part using a 12.7 mm diameter twist drill with a point angle of 118 degrees. The ho
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Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answer:
maximum speed for safe vehicle operation = 55mph
Explanation:
Given data :
radius ( R ) = 678 ft
old building located ( m )= 30 ft
super elevation = 0.06
<u>Determine the maximum speed for safe vehicle operation </u>
firstly calculate the stopping sight distance
m = R ( 1 - cos ) ---- ( 1 )
R = 678
m ( horizontal sightline ) = 30 ft
back to equation 1
30 = 678 ( 1 - cos (28.655 *s / 678 ) )
( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044
cos = 1.044
hence ; 28.65 * s = 678 * 0.2956
s = 6.99 ≈ 7 ft
next we will calculate the design speed ( u ) using the formula below
S = 1.47 ut + ---- ( 2 )
t = reaction time, a = vehicle acceleration, G1 = grade percentage
assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0
back to equation 2
6.99 = 1.47 * u * 2.5 +
3.675 u + 0.0958 u^2 - 6.99 = 0
u ( 3.675 + 0.0958 u ) = 6.99
Answer:
(a) 0.12924
(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Explanation:
(a)
n=200 for fair coin getting head, p= 0.5
Expectation = np =200*0.5=100
Variance = np(1 - p) = 100(1-0.5)=100*0.5=50
Standard deviation,
Z value for 108,
P( x ≥108) = P( z >1.13)= 0.12924
(b)
Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.