Select all of the answers that apply.
2 answers:
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Answer:
Option D is the correct answer.
Explanation:
Stress is the force per unit area that tend to change the shape of body.
Stress is defined as internal resistive force per unit area.
So, so stress distributed over an area is best described as internal resistive force.
Option D is the correct answer.
Answer:
Vf = 29.4 m/s
h = 44.1 m
Explanation:
Data:
- Initial Velocity (Vo) = 0 m/s
- Gravity (g) = 9.8 m/s²
- Time (t) = 3 s
- Final Velocity (Vf) = ?
- Height (h) = ?
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Final Velocity
Use formula:
- Vf = g * t
Replace:
- Vf = 9.8 m/s² * 3s
Multiply:
- Vf = 29.4 m/s
==================================================================
Height
Use formula:
Replace:
Multiply time squared:
Simplify the s², and multiply in the numerator:
It divides:
What is the velocity when falling to the ground?
The final velocity is <u>29.4 meters per seconds.</u>
How high is the building?
The height of the building is <u>44.1 meters.</u>
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s