The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s
1 answer:
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Answer:
Explanation:
= Angular speed
= Distance of Mary = 11.5 ft
= Distance of Alex = 6 ft
Ratio of centripetal acceleration is given by
Mary's centripetal acceleration is 1.92 times the centripetal acceleration of Alex
Answer:
<em>The force required is 3,104 N</em>
Explanation:
<u>Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = ma
Where a is the acceleration of the object.
On the other hand, the equations of the Kinematics describe the motion of the object by the equation:
Where:
vf is the final speed
vo is the initial speed
a is the acceleration
t is the time
Solving for a:
We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:
The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:
F= 3,104 N
The force required is 3,104 N
Answer:
The pickup truck and hatchback will meet again at 440.896 m
Explanation:
Let us assume that both vehicles are at origin at the start means initial position is zero i.e. = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.
Truck:
= 33.2 m/s, a = 0 (since the velocity is constant),
= 0
Using
s = 33.2t .......... eq (1)
Hatchback:
,
= 0 m/s (since initial velocity is zero),
= 0
Using
putting in the data we will get
now putting 's' value from eq (1)
which will give,
t = 13.28 s
so both vehicles will meet up gain after 13.28 sec.
putting t = 13.28 in eq (1) will give
s = 440.896 m
So, both vehicles will meet up again at 440.896 m.