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vovangra [49]
3 years ago
15

A child is pushing a merry-go-round. the angle through which the merry-go-round has turned varies with time according to θ(t)=γt

+βt3, where γ= 0.422 rad/s and β= 1.35×10−2 rad/s3.

Physics
1 answer:
Andreyy893 years ago
3 0
I attached the missing part of the question.
Part A
Angular velocity is simply the rate at which angle changes, in other words, it is the first derivative of the angle function with respect to time:
\theta (t)=\gamma t+\beta t^3\\ w(t)=\frac{d\theta(t)}{dt}\\ w(t)=\gamma +3\beta t^2
Part B
Initial value is the value at t=0. To find initial value we simply plug t=0 into the equation.
w(0)=\gamma +3\beta\cdot 0^2\\
w(0)=\gamma\\
w_0=\gamma= 0.422\frac{rad}{s}
Part C
To find these values we simply need to plug in t=5 in the equation.
w(5)=\gamma +3\beta\cdot 5^2\\
w(5)=\gamma +75\cdot \beta=0.422+75 \cdot 1.35\cdot10^{-2}=1.43\frac{rad}{s}
Part D
Average angular velocity is total angular displacement divided by time:
w_{av}=\frac{\Delta \theta}{\Delta t}
\Delta \theta=\theta(5)-\theta(0)=\gamma\cdot5+\beta 5^3\\
\Delta \theta=0.422\cdot5+125\cdot1.35\cdot10^{-2}=3.7975 $rad
The average angular velocity is:
w_{av}=\frac{\Delta \theta}{\Delta t}=\frac{3.7975 }{5}=0.7595\frac{rad}{s}

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Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
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Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

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Answer:

(9.64 +- 0.86) m/s^2

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Where

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If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:

x = \frac{1}{2}*a * t^2

And the acceleration can be obtained as:

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Where x is the distance fallen and a = g.

So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate

g = 2*\frac{100}{144^2} = 9.64e-3 \frac{mm}{ms^2} = 9.64 \frac{m}{s^2}

For the uncertainty we have to calculate the relative uncertainties first

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For multiplications or divisions the relative uncertainties are added

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We convert this into absolute uncertainty:

(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)

Finally, this is multiplied by a constant scalar, so:

2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)

We convert the units

0.86 m/(s^2)

And the measurement is (9.64 +- 0.86) m/s^2

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