||Homework Help||Kindergarten-University. - Page 73

Created

Last reply

Replies

739

Views

160674

Users

76

Likes

427

Frequent Posters

akhl thumbnail
Anniversary 17 Thumbnail Group Promotion 5 Thumbnail Fascinator 1 Thumbnail
Posted: 9 years ago
Originally posted by: Roses-khushi

hello i need helpp

give the structural formulas of

6- 1-chloro-2-methyl-2-butene

7- 3-ethyl-1-heptyne


6)

Cl
 |
CH2 - C = CH - CH3
           |
          CH3



7)

CH triple C - CH - CH2 - CH2 - CH2 - CH3
                     |
                    CH2
                     \
                     CH3


Chemcart_MJ thumbnail
Anniversary 13 Thumbnail Group Promotion 8 Thumbnail + 5
Posted: 9 years ago
can any1 help me with this question
A demonstration setup consists of a uniform board of length L, hinged at the bottom end and elevated at an angle by means of a support stick. A ball rests at the elevated end, and a light cup is attached to the board at the distance d from the elevated end to catch the ball when the stick supporting the board is suddenly removed. You want to use a thin hinged board 1.00 m long and 10.0 cm wide, and you plan to have the vertical support stick located right at its elevated end 
A) How long can you make the support stick maximally so that the ball has a chance to be caught?
Assume that you choose to use the longest possible support stick placed at the elevated end of the board. What distance d from that end should the cup be located to ensure that the ball will be caught in the cup?

akhl thumbnail
Anniversary 17 Thumbnail Group Promotion 5 Thumbnail Fascinator 1 Thumbnail
Posted: 9 years ago
Originally posted by: _Meena_

can any1 help me with this question

A demonstration setup consists of a uniform board of length L, hinged at the bottom end and elevated at an angle by means of a support stick. A ball rests at the elevated end, and a light cup is attached to the board at the distance d from the elevated end to catch the ball when the stick supporting the board is suddenly removed. You want to use a thin hinged board 1.00 m long and 10.0 cm wide, and you plan to have the vertical support stick located right at its elevated end 
A) How long can you make the support stick maximally so that the ball has a chance to be caught?
Assume that you choose to use the longest possible support stick placed at the elevated end of the board. What distance d from that end should the cup be located to ensure that the ball will be caught in the cup?


The cup will catch the ball if the following two conditions are satisfied: -
1) The board falls faster than the ball.
2) When the entire board touches the ground, the distance of the ball from the hinge = the distance of the cup from the hinge

Consider condition (1).
The ball is at the edge of the board.
Let tangential acceleration of the edge of the board = a
The vertical component of this tangential acceleration, ay = a cos(theta)
The acceleration of the ball = g, vertically downward

Condition (1) is satisfied if 
ay > g

Or, a cos(theta) > g

But a = L * alpha, where alpha = angular acceleration of the board
Therefore,
L alpha cos(theta) > g ------------------ (1)

alpha = tau/I ------------------------------ (2)
(Here tau = torque on the board and I = moment of inertia of the board)

The torque of any force passing through the axis = 0
The only other force on the board = the force of gravity = Mg
The force of gravity acts vertically downward at a distance L/2 from the axis

Therefore, tau = Mg (L/2) cos(theta) -------------------- (3)
I = (1/3) M L^2 ---------------------------- (4)

Using equations (3) and (4) in equation (2),
alpha = (3g/2L)  cos(theta)

Substituting the value of alpha in the inequality (1)
L (3g/2L) cos^2 (theta) > g

Dividing both sides by g,
(3/2) cos^2 (theta) > 1
Or, cos^2 (theta) > 2/3
Or, cos (theta) > sqrt(2/3)

Minimum value of cos (theta) = sqrt(2/3) ----------- (5)
Therefore maximum value of sin(theta) = sqrt(1  -  2/3) = 1/sqrt(3) ------- (6)

H = L sin(theta)
When sin(theta) is maximum, then H is maximum
Therefore maximum value of H = L/sqrt(3) = 1.00 m/sqrt(3) = 0.577 m

Consider condition (2).
The cup is attached to the board, so its distance from the hinge does not change but the ball falls vertically downward.

When the entire board reaches the ground: -
The horizontal distance of the ball from the hinge = L cos(theta), and
the horizontal distance of the cup from the hinge = L - d

Condition (2) is satisfied if 
L cos(theta) = L - d
Or, d = L - L cos(theta)
Or, d = L (1 - cos theta)

Substituting the value of cos theta from equation (5),
d = L (1 - sqrt(2/3))
Or, d = L (0.184 = 1.00 m * 0.184 = 0.184 m

Answer: Maximum length of the support stick = L/sqrt(3) = 0.577 m
d = L (1 - sqrt(2/3)) = 0.184 m


Edited: The previous answer was also perfectly correct. Edited to make it more clear.

Edited by akhl - 9 years ago
Chemcart_MJ thumbnail
Anniversary 13 Thumbnail Group Promotion 8 Thumbnail + 5
Posted: 9 years ago
how do i solve this polymer?
2-Methoxypropene will form a polymer when treated with a radical initiator. Write the structure of this polymer (draw only three monomer lengths of this polymer.)
akhl thumbnail
Anniversary 17 Thumbnail Group Promotion 5 Thumbnail Fascinator 1 Thumbnail
Posted: 9 years ago
Originally posted by: _Meena_

how do i solve this polymer?

2-Methoxypropene will form a polymer when treated with a radical initiator. Write the structure of this polymer (draw only three monomer lengths of this polymer.)


2-methoxypropene is 

            CH3
          /
         O
          |
CH2=C-CH3

The polymer is
                 CH3                   CH3                  CH3
                /                         /                        /
               O                      O                      O
                |                       |                        |
... -CH = C - CH2 - CH = C - CH2 - CH = C - CH2- ...

Make sure to put dots in beginning and end to show that the unit keeps repeating.
Edited by akhl - 9 years ago
Chemcart_MJ thumbnail
Anniversary 13 Thumbnail Group Promotion 8 Thumbnail + 5
Posted: 9 years ago
Originally posted by: akhl


2-methoxypropene is 

            CH3
          /
         O
          |
CH2=C-CH3

The polymer is
                 CH3                   CH3                  CH3
                /                         /                        /
               O                      O                      O
                |                       |                        |
... -CH = C - CH2 - CH = C - CH2 - CH = C - CH2- ...

Make sure to put dots in beginning and end to show that the unit keeps repeating.

i did get similar structure but i'm exactly confused on what they mean by "draw only three monomer lengths of this polymer"
akhl thumbnail
Anniversary 17 Thumbnail Group Promotion 5 Thumbnail Fascinator 1 Thumbnail
Posted: 9 years ago
[quote=_Meena_]i did get similar structure but i'm exactly confused on what they mean by "draw only three monomer lengths of this polymer"[/quote]
As you can see, I have shown 3 monomers in the answer.

Bobbi. thumbnail
Anniversary 14 Thumbnail Group Promotion 6 Thumbnail + 4
Posted: 9 years ago
Anyone here studying English law?
akhl thumbnail
Anniversary 17 Thumbnail Group Promotion 5 Thumbnail Fascinator 1 Thumbnail
Posted: 9 years ago
Can you point me to any recent (within 5 years) paper or patent, which explains how to repurpose some other sensor into a bio-medical sensor? The other sensor could be environmental, military, chemical, or food sensor.
Chemcart_MJ thumbnail
Anniversary 13 Thumbnail Group Promotion 8 Thumbnail + 5
Posted: 9 years ago
can anyone help me in these 2 questions