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Keith_Richards [23]
3 years ago
14

A pendulum can be formed by tying a small object, like a tennis ball, to a string, and then connecting the other end of the stri

ng to the ceiling. Suppose the pendulum is pulled to one side and released at t1. At t^2, the pendulum has swung halfway back to a vertical position. At t^3, the pendulum has swung all the way back to a vertical position. Rank the three instants in time by the magnitude of the centripetal acceleration, from greatest to least. Most of the homework activities will be Context-rich Problems.
Physics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

1- t^3

2- t^2

3- t1

Explanation:

The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:

ac = v²/r

where,

ac = centripetal acceleration

v = speed

r = radius

for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:

<u>1- t^3</u>

<u>2- t^2</u>

<u>3- t1</u>

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In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
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4 0
3 years ago
A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acce
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force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

8 0
2 years ago
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Explanation:

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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
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Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

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Answer:

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m = mass of the block

Maximum acceleration of the block is given as

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f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72

8 0
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