Explanation:
Value of the cross-sectional area is as follows.
A =
= 3.45 
The given data is as follows.
Allowable stress = 14,500 psi
Shear stress = 7100 psi
Now, we will calculate maximum load from allowable stress as follows.

= 
= 50025 lb
Now, maximum load from shear stress is as follows.

= 
= 48990 lb
Hence,
will be calculated as follows.

= 48990 lb
Thus, we can conclude that the maximum permissible load
is 48990 lb.
Answer:
Yes it is
Explanation:
the sum moles at the left side equals the sum of moles at the right side
Answer:
T = 1.2 s
T = 15.1 m = 15 m
Explanation:
This is a case of projectile motion:
TOTAL TIME OF FLIGHT:
The formula for total time of flight in projectile motion is:
T = 2 V₀ Sinθ/g
where,
T = Total Time of Flight = ?
V₀ = Launch Speed = 13.9 m/s
θ = Launch Angle = 25°
g = 9.8 m/s²
Therefore,
T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)
<u>T = 1.2 s</u>
<u></u>
RANGE OF BALL:
The formula for range in projectile motion is:
R = V₀² Sin2θ/g
where,
R = Horizontal Distance Covered by ball = ?
Therefore,
T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)
<u>T = 15.1 m = 15 m</u>