Answer:
a) 3.5 m
b) 14 secs
c) 1.4 secs
Explanation:
<u>a) Determine the distance the particle will travel</u>
given velocity ( final velocity ) = 5 m/s
v^2 = u^2 + 2as
s = ( v^2 - u^2 ) / 2a
= ( 5^2 - 8^2 ) / 2 ( -0.5 * 5^3/2 )
= 3.5 m
<u>b) Determine the time when v = 1m/s</u>
V = u + at
1 = 8 + ( -0.5 * 1^3/2 ) * t
∴ t = 14 secs
c) Determine the time required for particle to travel 8 m
<em>we will employ both equations above </em>
V^2 = u^2 + 2as
s = 8 m , V = unknown , u = 8 m/s back to equation
V^2 = 8^2 + 2 ( - 1/2 * V^3/2 ) * 8
∴ V^2 + 8V^3/2 - 64 = 0
resolving the above equation
V = 3.478 m/s
now using the second equation
V = u + at
3.478 = 8 + ( - 1/2 * 3.478^3/2 ) * t
hence : t = 1.4 secs
Answer:
Circular tube
Explanation:
Now for better understanding lets take an example
Lets take
Diameter of solid bar= 
cm
Outer diameter of tube =6 cm
Inner diameter of tube=2 cm
So from we can say that both tubes have equal cross sectional area.
We know that buckling load is given as 
If area moment of inertia(I) is high then buckling load will be high.
We know that area moment of inertia(I)
For circular tube 
For circular bar 
Now by putting the values
For circular tube 
For circular bar 
So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.
Answer:
C: Stop before entering the pedestrian crosswalk.
Explanation:
Answer:
c
Explanation:
This is because many things, such as pcs, over heat
Answer:
Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]
Explanation:
The execution in the option D is correct. This is because there is more than one reasonable criterion.
