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11Alexandr11 [23.1K]
4 years ago
6

How long must a simple pendulum be if it is to make exactly one swing per second? (That is, one complete vibration takes exactly

2.0 s.)
Physics
1 answer:
umka2103 [35]4 years ago
8 0

Oscillation is a type of periodic motion which repeats itself to and from about a point which is called mean position. The period of a Pendulum can be described as a ratio between the length and gravity as,

T = 2\pi \sqrt{\frac{L}{g}}

Here,

L = length

g = Gravity

If we rearrange  to find the length we have that,

L = \frac{T^2g}{4\pi^2}

Our period is 2s and the gravity is 9.8m/s^2, then,

L = \frac{(2)^2(9.8)}{4\pi^2}

L = 0.9929m

The simple required length of the pendulum must be 0.9929m

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A car drives around a racetrack for 30 seconds. What do you need to know to
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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
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Answer:

h = 3.5 m

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2gh = v_f^2 - v_i^2\\

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g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

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m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

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v₁ = vf = final velocity of ball while collision = 8.3 m/s

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Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

6 0
3 years ago
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