We know, Period = 1/frequency = 1/20 sec.
So, your final answer is 0.05 sec
Hope this helps!
<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
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The answer would be 981 newtons or 220.46 pounds.
The normal stress follows the formula written below:
σ = F/A
There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,
A = πr² = π(0.75 in)² = 1.767 in²
So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>
Answer:
The tension is
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is
The mass tied between the two cloth line is
The distance it sags is
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as
To obtain we apply SOHCAHTOH Rule
So