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.
Air resistance force
tension force
spring force
frictional force
normal force
gravitational force
applied force
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Answer:
(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced
(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained
Explanation:
(a)
To find the final velocity
for an object traveling distance h taking the initial vertical component of velocity as
the kinematics equation is written as
where a is acceleration
Substituting g for a where g is gravitational force value taken as 9.81

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h
= 20.44275
Therefore, the divers enter with a speed of 20.4 m/s
The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced
(b)
The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

Since we have final velocity of 25 m/s


= 14.390761 m/s
Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s
In conclusion, the upward initial velocity can’t be physically attained
The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
<h3>
What is average speed?</h3>
The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.
<h3>Total time taken by the car during the entire race</h3>
time = distance/average speed
time = (1.41 km) / (278 km/h)
time = 0.0051 hr
The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;
d = 705 m = 0.705 km
t1 = 0.705/210
t1 = 0.0034 hr
<h3>time for the second half</h3>
t2 = 0.0051 - 0.0034 hr
t2 = 0.0017 hr
<h3>minimum average speed of the second half</h3>
v = d/t
v = 0.705 km / 0.0017 hr
v = 414.7 km/hr
Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
Learn more about average speed here: brainly.com/question/4931057
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