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Brrunno [24]
2 years ago
7

A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillat

es with amplitude A = 1 cm and period T = 2 seconds. Which of the following alterations will change the period of the pendulum’s oscillations to T ′ = 1 second? 1. Using a lighter brass sphere of mass m ′ ≈ 0.25 kg. 2. Using a shorter string of length L ′ ≈ 0.25 meters. 3. Using a longer string of length L ′ ≈ 2 meters. 4. Using a longer string of length L ′ ≈ 4 meters. 5. Using a shorter string of length L ′ ≈ 0.5 meters. 6. Increasing the oscillation amplitude to A ′ = 4 cm. 7. Using a heavier brass sphere of mass m ′ ≈ 4 kg. 8. Reducing the oscillation amplitude to A ′ = 0.25 cm.
Physics
1 answer:
hram777 [196]2 years ago
6 0

Answer:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

Explanation:

The period of a pendulum is given by

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

where

L is the length of the pendulum

g is the acceleration due to gravity

We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.

Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.

Therefore, the only alterations that will reduce the period of the pendulum are:

2. Using a shorter string of length L ′ ≈ 0.25 meters

5. Using a shorter string of length L ′ ≈ 0.5 meters

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AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
2 years ago
You are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform m
mamaluj [8]
The correct answer is:
<span>paramagnetism 

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A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
Serggg [28]

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

L=I\omega

where

I is the moment of inertia

\omega is the angular speed

Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

\omega_1 = 5.00 rad/s is the initial angular speed

I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

7 0
2 years ago
As the drawing illustrates, a siren can be made by blowing a jet of air through 20 equally spaced holes in a rotating disk. The
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Answer:

ω = 630.2663 = 630[rad/s]

Explanation:

Solution:

- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.

                                  1rev(20 hole) -> 20(cycle)/rev  

                                        2006.2(cycle) -> f ?  

                              f = 2006.2/20 = 100.31rev at second  

- The relation between angular frequency and angular speed is given by:

                                 ω = 2πf

                                 ω = 2*3.14*100.31

                                 ω = 630.2663 = 630[rad/s]

4 0
2 years ago
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Lapatulllka [165]

Answer:

X= 700 Joules

Explanation:

The question asked about the efficiency of the work done.

The formula for efficiency is: Efficiency = (Useful output / input work) * 100%

The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.

Using the formula of efficiency,

20 = (140/X) * 100

So, we simply solve the above equation.

X= 140*100/20

X= 700 Joules

6 0
2 years ago
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