bachelor's degree in mechanical engineering or mechanical engineering technology
Answer:
R = 148.346 N
M₀ = - 237.2792 N-m
Explanation:
Point O is selected as a convenient reference point for the force-couple system which is to represent the given system
We can apply
∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N
∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N
Then
R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)
⇒ R = 148.346 N
Now, we obtain the moment about the origin as follows
M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)
We can see the pic shown in order to understand the question.
Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:

R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

c) By using the equation of the ideal gases you obtain:

In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.
<h3>What is a
thrust angle?</h3>
A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.
This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.
Read more on thrust angle here: brainly.com/question/13000914
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I think it’s is false I’m not that sure