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timurjin [86]
3 years ago
8

A pendulum is hanging from a point and its total mechanical energy is 20,000 J. Neglecting friction, if energy is conserved, wha

t values should you put in the blanks.
1. At which point/s in the pendulum motion do you think is most difficult to stop?


2. At which point/s would the pendulum easiest to stop?


3. If friction were not present, how much total mechanical energy would the pendulum have at:


a. point A? _______ d. point D? ______


b. point B? _______ e. point E? ______


c. point C? _______
Physics
1 answer:
AlexFokin [52]3 years ago
5 0

(1) The pendulum will have maximum velocity at the lowest point and it will be the most difficult point to stop.

(2) Velocity at maximum height is zero and it will be the easiest point to stop the pendulum.

(3) The total mechanical energy of the pendulum would  be the same at all point.

<h3>Point of maximum velocity of the pendulum</h3>

A pendulum has maximum kinetic energy at the lowest point. The kinetic energy is given as;

K.E = ¹/₂mv²

The pendulum will have maximum velocity due to maximum kinetic energy at the lowest point and it will be the most difficult point to stop.

<h3>Point of maximum height of the pendulum</h3>

A pendulum has maximum potential energy at the highest point. The potential energy is given as;

P.E = mgh

Velocity at maximum height is zero and it will be the easiest point to stop the pendulum.

<h3>Conservation of energy</h3>

In absence of friction, the total mechanical energy will be conserved. Thus, the total mechanical energy of the pendulum would  be the same at all point.

Learn more about pendulum here:brainly.com/question/26449711

#SPJ1

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GarryVolchara [31]

Answer:

Imp = 11.666\,\frac{kg\cdot m}{s}

Explanation:

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(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}

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After collision:

\frac{1}{2}\cdot (0.320\,kg)\cdot v_{B}^{2} = (0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (15\,m)

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The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

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5 0
3 years ago
Read 2 more answers
Insert
nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

A\propto N

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

A\propto m

where m is the mass of the sample.

In this problem:

- When the mass is m_1 = 1 g, the activity is A_1=16 count/min

- When the mass is m_2=20 g, the activity is A_2

So we can find A2 by using the rule of three:

\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min

B)

The equation describing the activity of a radioactive sample as a function of time is:

A(t)= A_0 e^{-\lambda t} (1)

where

A_0 is the initial activity at time t = 0

t is the time

\lambda is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

\lambda = \frac{ln2}{t_{1/2}}

where t_{1/2} is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

t_{1/2}=5700 y

So its decay constant is

\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

A_0 = 320 count/min (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min

C)

We know that a sample of living wood has an activity of

A=16 count/min per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

A_0 = A\cdot 5 = (16)(5)=80 count/min

Moreover, here we have a sample of 5 g, with current activity of A=20 count/min: it means that its activity per gram of mass is

A'=\frac{20}{5}=4 count/min

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

t=2t_{1/2}=2(5700)=11,400 y

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