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PHY101 - Physics Assignment No. 02 Solution and Discussion Fall 2013 Due Date: 09 /1 2 /2013

Assignment 2: (Fall 2013)
PHYSICS (PHY101)
TOTAL MARKS: 35
Due Date: 09/12/2013
DON’T MISS THESE Important instructions:
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Don’t wait for the last date to submit your assignment.
Question # 1
Find out the magnitude and direction of the overall acceleration in the fig shown? Marks = 6
Question # 2
In its natural state, an average of 5.5×106Kg of water flowed per second over Niagara Falls, falling 55m. If all the work done by gravity could be converted into electric power as the water fall to the bottom, how much power would the falls generate? Marks = 8
Question # 3
The solid uniform disk rolls without slipping. What is its kinetic energy as shown figure? Marks =8
Question # 4
How much work is done when the wheel rotates shown in the figure? Marks = 6
Question # 5
The tunable shown above makes a record rotate at 33.3 revolutions per minute. If the linear speed of an ant located at radius r is v, what is the linear speed of a point on the outer edge of the record at radius 3r? Marks = 3+4
a. also v
b. 1\3v
c. 3v
d. 9v
e. 6v
Write the reason of your above selected choice as well.
…………Good Luck…………>>>>>>>>>>>>

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please share your ideas

Please Discuss here about this assignment.Thanks

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Answer of question no.3 is

A rolling disk will have two forms of KE, 
so that the total KE will be the sum of its linear and angular KE's. 

K.E=angular K.E + linear K.E
r = 0.356 m 
m = 3.50 kg 
w = 1.55 rad/s 

I = 1/2mr²

 = (0.5)(3.50)(0.356)²

= 0.222 kg-m² 
Angular KE of rolling disk = 1/2Iw²

= (0.5)(0.222)(1.55)²

= 0.267 J 
Linear KE of rolling disk = 1/2mV²

 = 1/2m(rw)²

= (0.5)(3.50)[(0.356)(1.55)]²

 = 0.533 J 
Total KE of rolling disk

Total K.E=Angular K.E + Linear K.E

 = 0.267 + 0.533

= 0.800 J

Answer of question no.4 is

W=?

τ=375N.m

θ=70degree

θ = 70(π/180 )              hint ( π/180 =degree)
θ = 1.22 radians 
 W = τθ 
W = 375 N.m(1.22 radians) 

W = 457.5 J

Answer of question no. 5 is

Velocity is the angular velocity times the radius. 

v = ωr 

as the angular velocity is constant throughout the whole disc, velocity is directly proportional to r 

v  ∝ r

3v = ω(3r) 

So the correct option is 3v.

Answer of question no.2 is

P=?

P=W/t

m=5.5×106Kg

d=h=55m

t=1sec

W=Potential Energy = mgh                 Hint=work done is converted in electric energy

W=(5.5×106Kg)(9.8)(55)

W=2964.5×106J

P=W/t

P=2964.5×106/1

P=2964.5×106w

tarik bhai q2 me apny 5.5 ko maltiply nhe kiya

W=(5.5×106Kg)(9.8)(55)         r agr 5.5 se maltiply krn tu 163047.5 *10 power 6 answer ata hy kiya ye tik hy ya jo apny kiya wo??????plzzz hlep me.....thanks

 

Hi, Sir 

         m = 5.5*106 kg

         This wrong value

 This is 5.5*10^6

                 10 power 6

2964.5 *10 pow 6

I enter the value in potential energy is 2964.5 × 10 power 6 .Power is not working in replay but u enter the value 10 power 6 it is not 106 

Answer of question 1 is   

Centripetal acceleration ( a c )=0.75 m/s2

tangential acceleration ( a t )=0.45 m/s2

atotal=√ac2+at2

atotal =√(.75)2+(.45)2

atotal=√(.5625)+(.2025)

atotal=√.765

atotal=.874 m/s2

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