A good engineer will create an outline that meets the useful necessities of the venture and after that play out an anxiety examination to get certainty that the tower will be sheltered under the normal scope of working conditions. By then, the outline will be tastefully satisfying to engineers. On the off chance that the customer needs something all the more stylishly cleaned, he may need to utilize a plan authority. A designer who puts feel initially is taking a chance with his expert standing.
Answer:
82 degrees
Explanation:
consider your staying point to be the center of a circle. this center has the coordinates (0, 0).
the radius of the circle is the distance you walked East (14 miles).
I assume your teacher means as "angle of displacement" the angle between the East-West line going through your starting point and the direct line from your starting point to your current position.
then the 100 miles North is tan(displacement angle)×14.
as it is the same, if you first went North and then East, or the other way around. you end up at the same point, with the same coordinates.
so, again.
100 = 14×tan(angle)
tan(angle) = 100/14 = 50/7 = 7.142857...
the displacement angle is then 82 degrees.
Answer:
A,B,C,D, and F are correct
Explanation:
Answer:
I = 0.451 amp
Explanation:
given,
C = 8.0 µF
V = 2 V
resistor connected between two terminal = 6 Ω
current flowing through resistor = 13 µsec
Q = CV
Q = 8 x 2
Q = 16 µC
for an RC discharge circuit
V = V_0e^{-\dfrac{t}{RC}}
I = \dfrac{-Q_0}{RC}e^{-\dfrac{t}{RC}}
t = 13 µsec
I = \dfrac{-16}{6\times 2}e^{-\dfrac{13}{6\times 2}}
I = 0.451 amp
neglecting -ve sign just to show direction.
Answer:
The observer could see to a depth of 4.38 m
Explanation:
Please check attachment for diagram.
Mathematically, from Snell law;
n1sin theta = n2 sin theta
1 * sin 90 = n2 * sin θR
where n2 = 1.33
1/1.33 = sin θR
Sin θR = 0.7519
θR = arc sin 0.7519
θR = 48.76
Now to get the height, we use the triangle
Using trigonometric ratio;
Tan( 90- θR) = H/5
H = 5 Tan( 90 - θR)
H = 5 Tan( 90-48.76)
H = 5 Tan41.24
H = 4.38 m
