logo

Crowdly

Browser

Додати до Chrome

PHS1011 - Classical physics and relativity - S1 2025

Шукаєте відповіді та рішення тестів для PHS1011 - Classical physics and relativity - S1 2025? Перегляньте нашу велику колекцію перевірених відповідей для PHS1011 - Classical physics and relativity - S1 2025 в learning.monash.edu.

Отримайте миттєвий доступ до точних відповідей та детальних пояснень для питань вашого курсу. Наша платформа, створена спільнотою, допомагає студентам досягати успіху!

Consider the following constant acceleration equations for rotational motion

 

Rearrange these equations to calculate at time t (the time at which the block has completed one full rotation, .

0%
0%
0%
0%
0%
Переглянути це питання

If is defined to be the mass of the block, then from question 3, we know that

gravitational PE lost by block-Earth = linear KE gained by block + rotational KE gained by disc

Or,

 

Substitute your answers to the previous questions into this equation and rearrange to find . Select the correct answer from the multiple choices given below.

Переглянути це питання

From the last question, we have that

 

Derive an expression for the uncertainty on in terms of the uncertainty on and

Hint, you may want to check over Section 3.3 of the Guide to experimental work.

 

Переглянути це питання

The disc is released from rest. The falling block causing it to rotate. After a time t the block has fallen a height h and the disc has rotated through an angle θ=2 (i.e. the disc has completed one full rotation). Assume that the string is long enough that the block still causes the disc to accelerate. 

If we define

  • : the height through which the block has fallen after time t
  • : the velocity of the block at time t
  • : the angular velocity of the disc at time t

Which of the following are the correct expressions for and in terms of , the radius of the hub?

Переглянути це питання

Sometimes, we can't measure the moment of inertia directly. For example if the object is oddly shaped. In such cases, we can use an indirect way to measure the moment of inertia - this is described in the Introduction of the lab book here (we recommend you read/watch this before answering the questions).

The general principle is that a falling block causes the disc to rotate, such that at any time

gravitational PE lost by block-earth =     linear KE gained by block        +    rotational KE gained by disc

If is the moment of inertia of the disc and is its angular velocity, which expression allows you to calculate the rotational energy of the disc?

Переглянути це питання

We record the following data:

Mass of disc, md = 0.800 ± 0.005 kg 

Radius of disc, r = 0.215 ± 0.001 m

Calculate the measured moment of inertia using the answer to the previous question. What is our uncertainty on our measured moment of inertia?

0%
0%
0%
0%
0%
Переглянути це питання

In your experiment, you will be measuring the moment of inertia of a solid disc using two different methods. In the first, direct method, we measure the mass (md) of the disc and its radius (r). Which equation gives the resulting moment of inertia of the disc about its centre?

 

0%
0%
0%
0%
Переглянути це питання

Which of the following would help most in an effort to obtain tighter constraints on the fit parameters?

Переглянути це питання

If the impulse momentum theorem holds, what do we expect the slope to be when fitting Change in momentum of the cart vs Impulse?

Enter the value of the expected slope which should be dimensionless (just the numerical value). 

Переглянути це питання

If the impulse momentum theorem holds, what do we expect the intercept to be when fitting Change in momentum of the cart vs Impulse?

Enter the value of the expected intercept in kg m/s but do not enter the units (just the numerical value, i.e., for an expected intercept of 3 Ns you would enter 3). 

Переглянути це питання

Хочете миттєвий доступ до всіх перевірених відповідей на learning.monash.edu?

Отримайте необмежений доступ до відповідей на екзаменаційні питання - встановіть розширення Crowdly зараз!

Browser

Додати до Chrome