Question 1
2 answers:
<u><em>Answer:</em></u>
A) Gravity.
<em><u>Explanation:</u></em>
Earth is stuck in orbit around the sun, because of the Sun's gravitational pull. In this scenario, the string represents gravity, keeping the Earth attracted towards the sun.
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Answer:
Refer to the attachment for solution (1).
<h3><u>Calculating time taken by it to stop (t) :</u></h3>By using the second equation of motion,
→ v = u + at
- v denotes final velocity
- u denotes initial velocity
- t denotes time
- a denotes acceleration
→ 0 = 5 + (-5/6)t
→ 0 = 5 - (5/6)t
→ 0 + (5/6)t = 5
→ (5/6)t = 5
→ t = 5 ÷ (5/6)
→ t = 5 × (6/5)
→ t = 6 seconds
→ Time taken to stop = 6 seconds
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