Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
The law of energy conservation states that energy cannot be created or destroyed, so choices B to D are immediately invalid. Choice A can explain this occurrence: <u>A. Some of the energy is used to combat friction, and thus is transformed from mechanical energy to heat.</u>
<h3>With a uniform acceleration of 2 m/s ²</h3>
First we need to convert the mm to inches to make our computation
easier.
1mm = 0.0393701
32mm * 0.0393701 = 1.25 in
Solution:
C = 1/2d = ½ (1.25) = 0.625 in^4
Tension: tension = Te/J = 2T/ piC^3
= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi
Bending:
I = pi/4 * c^4 = 119.842 x 10^-3 in^4
M = (5)(600) = 3600 lb in
G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2
psi = -18.775ksi
Gx = -18.775 ksi
Gy = 0
Txy = 6.519 ksi
G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi
1.
G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi
G2 = Gave - R = -9/387 - 11.429 = -20.8
Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 =
-1.3889
ϴp = -27.1 degrees and 62.9 degrees
2.
Tmax = R = 11.43 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2
+ (6.519)^2 = 11.429 ksi
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