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Roses Khushi

@Roses-khushi

Sparkler

+ 2

Posted: 10 years ago

#691

hello i need helpp

give the structural formulas of

1- 2,2,3,3, tetramethyl pentane

2- 1,2, dibromo-2-methyl propane

3- diethyl Acetylene

4- 3-bromo-4-methyl-3-hexene

5- 3,3-dimethyl-1-butyne

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

7- 3-ethyl-1-heptyne

plzzz help its my H.W

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akhl

@akhl

Dazzler

Posted: 10 years ago

#692

Originally posted by: Roses-khushi
hello i need helpp

give the structural formulas of

1- 2,2,3,3, tetramethyl pentane

2- 1,2, dibromo-2-methyl propane

3- diethyl Acetylene

4- 3-bromo-4-methyl-3-hexene

5- 3,3-dimethyl-1-butyne

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

7- 3-ethyl-1-heptyne

plzzz help its my H.W

I can answer all of these but how do I share the images? If u PM me ur email id, i will send in that. Or do I put the images in any website?

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akhl

@akhl

Dazzler

Posted: 10 years ago

#693

Originally posted by: Roses-khushi
hello i need helpp

give the structural formulas of

1- 2,2,3,3, tetramethyl pentane

2- 1,2, dibromo-2-methyl propane

Edited by akhl - 10 years ago

Quote

akhl

@akhl

Dazzler

Posted: 10 years ago

#694

Originally posted by: Roses-khushi
hello i need helpp

give the structural formulas of

3- diethyl Acetylene

CH3 - CH2 - C triple - CCH2 - CH3

Note: Replace the word "triple" by triple bomd

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akhl

@akhl

Dazzler

Posted: 10 years ago

#695

Originally posted by: Roses-khushi

4- 3-bromo-4-methyl-3-hexene
5- 3,3-dimethyl-1-butyne

CH3 -CH -C=CH - CH2 - CH3

CH3

CH triple C - C - CH3

CH3

Replace the word "triple" by triple bond

Edited by akhl - 10 years ago

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akhl

@akhl

Dazzler

Posted: 10 years ago

#696

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

CH2 - C = CH - CH3

CH3

CH triple C - CH - CH2 - CH2 - CH2 - CH3

CH2

CH3

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MJ1009cc

@MJ_1009

Achiever

+ 7

Posted: 10 years ago

#697

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?

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akhl

@akhl

Dazzler

Posted: 10 years ago

#698

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 - 10 years ago

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MJ1009cc

@MJ_1009

Achiever

+ 7

Posted: 10 years ago

#699

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.)

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akhl

@akhl

Dazzler

Posted: 10 years ago

#700

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

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 - 10 years ago