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Old May 1st 2008, 09:47 PM   #1
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5 Question from Practice Exam, Please Help Me!

The Practice Exam I was given contains answer, but getting the answer is the hard part.

1. A rotating wheel requires 3.00 s to rotate 37 revolutions. At the end of 3.00 s the angular velocity will be 98.0 rad/s. What is the constant angular acceleration? (1 rev = 2pi rad)

Ans:13.7 rad/s^2


Im using the formula delta(theta) = (omega)(time) + 1/2at^2
delta(theta)= (37*2pi)
(omega)(time)= 98.0 rad/s * 3sec
1/2at^2= 1/2a(3sec)^2
I plug everything in, but i get -13.7rad/s^2. Am I just lucky I came close to negative answer or did i so something wrong?

2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams?

Ans: 139 gm

Yea, I'm clueless


3. Two spheres of equal volume 100 cm3 contain an inert gas at 0C and 1.00 atm pressure. They are joined by a small tube of negligible volume, allowing the gas to flow between the spheres. What common pressure will exist in the two spheres if one is raised to 100C while the other is kept at 0 C? Assume an ideal gas.

Ans: 1.16 atm

I try every way possible to manipulate Pv=nRT, but I can't get the answer

[s]4.A 5.00g bullet moving with an initial speed of 400 m/s is fired horizontally along the positive x-axis, passing completely through a 1.00-kg block, which initially is at rest on a frictionless horizontal surface at x = 0.00 cm. The block is connected to a spring having a force constant of 900 N/m. If the block moves 5.00 cm to the right (x = 5.00 cm) along the positive x-axis after impact, what is the speed of the bullet when it emerges from the block?[/s]

Ans:100 m/s

From the formual (KE+PEg+PEs)i = (KE+PEg+PEs)f
I'm got the formula: 1/2mv^2 = 1/2mv^2 + 1/2Kx^2
I don't know what I am doing wrong.


I just solved this

5. A thin spherical shell of mass 0.400 kg and diameter 0.200 m is filled with alcohol (rho = 806 kg/m3). It is released from rest at the bottom of a tank of water (rho = 1000 kg/m3). What is the acceleration of the alcohol spherical shell as it starts to rise to the surface?

Ans:1.06 m/s^2


Yea, another clueless one

Thank you, for whoever helps me on this

Last edited by Heavyarms2050; May 3rd 2008 at 05:05 PM.
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Old May 2nd 2008, 07:30 AM   #2
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Originally Posted by Heavyarms2050 View Post
2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams?

Ans: 139 gm

Yea, I'm clueless

I will help you with this question for now. Are you familiar with the principle of moments ?

If the meter stick is balanced at the 49.7cm mark, what does that tell you about the position of the centre of mass?

A good clear diagram always helps with a question like this, I have attached one.


first find the distance of the 10g block form the pivot.

39.2cm - 10cm = 29.2cm

and you need to find the distance of the centre of mass of the rules form the pivot.

49.7cm - 39.2cm = 10.5cm

using the principle of moment the anticlockwise moment causes by the mass is equal to the clockwise moment cause my the centre of mass about the pivot.

$\displaystyle 29.2cm \times 50 g = 10.5cm \times M $

solve the equation for M, you should get 139 grams.

Bobak
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Last edited by Bobak; May 2nd 2008 at 01:57 PM.
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Old May 2nd 2008, 01:54 PM   #3
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$\displaystyle
29.2cm \times 10 g = 10.5cm \times M
$

there was typo in your formula, it was suppose to be 50g not 10g. Thx a million though.

This is actually getting urgent now, but can you guys/anybody PLEASE help me on these 4 problems by Monday.
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Old May 3rd 2008, 12:18 PM   #4
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Originally Posted by Heavyarms2050 View Post
5. A thin spherical shell of mass 0.400 kg and diameter 0.200 m is filled with alcohol (rho = 806 kg/m3). It is released from rest at the bottom of a tank of water (rho = 1000 kg/m3). What is the acceleration of the alcohol spherical shell as it starts to rise to the surface?

Ans:1.06 m/s^2


Yea, another clueless one

I believe you have to use archimedes principle for this question. Do you know what it is?

I will quote it for you.

When a body is wholly or partly immersed in a fluid it experiences and upward force equal to the weight of the fluid displaced.

Firstly we must calculate the weight of the displaced water which. will be the volume of the sphere multiplied by the density of the water.

$\displaystyle Upthrust = 1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2 $

now you must draw a free body force diagram

note the mass of the sphere is $\displaystyle 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg $ this is the mass of the volume of the alcohol plus the mass of the shell




so the net force is $\displaystyle Upthrust - Weight $

using Newton's second law

$\displaystyle Mass \times Acceleration = Upthrust - Weight $
$\displaystyle ma= 1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2 - m \times 9.81 N$

so $\displaystyle a= \frac{1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2}{ 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg } - 9.81 m/s^2$

giving $\displaystyle a= 1.07195 m/s^2$
or $\displaystyle a= 1.06 m/s^2$ quoting to 3 significant figures.

I hope that helps

Bobak
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Last edited by Bobak; May 3rd 2008 at 03:31 PM.
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Old May 3rd 2008, 03:21 PM   #5
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$\displaystyle
a= \frac{1000 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 \times 9.81 m/s^2}{ 806 Kg /m^3 \times \frac{4}{3} \pi (0.1)^3 m^3 + 0.4 Kg } - 9.81 m/s^2
$


I'm not getting the answer using the formula you have given me for the 2nd

Last edited by Heavyarms2050; May 3rd 2008 at 10:07 PM.
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Old May 3rd 2008, 07:37 PM   #6
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Originally Posted by Heavyarms2050 View Post
2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams?

Ans: 139 gm
You neglected to mention that you are using $\displaystyle g = 10 m/s^2$. As most of the world does not use this (unrealistic) assumption you should always mention it when asking for help.

We know that the meter stick is balanced at the 49.7 cm mark, so this is the location of the CM.

I am going to take the sum of the torques about the 39.2 cm mark and call a CCW torque positive.

$\displaystyle \sum \tau = (0.0500~kg)(0.292~m) - Mg(0.105~m) = 0$
where M is the mass of the meter stick.

Solve for M.

-Dan
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