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samera12

@samera12

Explorer

Posted: 16 years ago

#121

thanks alot for ur time.. i will start working on it..😊

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Divya

@~Divya

Inactive

Posted: 16 years ago

#122

Hi Can you guys help me

Can you give me a Acrostic Poem For Perseverance

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maryam

@aishu_fan

Voyager

Posted: 16 years ago

#123

im confused about my phsycis. we are leranign the right hand rule to find the magnetic force on a proton and elctron using right hand rule. but they tell us to do that liek after we solve the equation.how r u supposed to do tht if the answer is like 5 N?

liek this question: if an electron in an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T west,what is the direction and magnitude of this velocity.

how do i FIND THE DIRECTION?ONLY DIRECTINO PLEASE

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akhl

@akhl

Dazzler

Posted: 16 years ago

#124

Note: If you consider direction of conventional current (Which is direction of positive charge and which is opposite to direction of electrons), then for the type of problem you have written, use left hand rule. Use right hand rule for induced current.

If you deal with electrons, then do just the opposite. In this problem we will deal with electrons and we will use right hand rule.

Make a fist out of your right hand. Then open your fore finger so that it points to the front. Open your thumb so that it points to the left. Open your middle finger and bend it so that it points down.

thumb points in the direction of force
fore finger points in the direction of field
middle finger points in the direction of velocity of electrons

Question says force is downward and field is west. Without changing relative orientations of your fingers, rotate your hand in such a way that thumb points down and forefinger points West. When you do that you will find that middle finger points north. This means electrons move to the north.
Ans: North

Instead of right hand, we could use left hand rule. That will give direction as south. But in left hand rule for such problems, middle finger is direction of conventional current. So conventional current is to the south. This means electron's velocity is to the north.

Do you know about unit vectors along coordinate axes and do you know about cross products between them? If yes, then I will tell you another method, which you may find easier.

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maryam

@aishu_fan

Voyager

Posted: 16 years ago

#125

thank u for ur help for the right hand rule!!i was wondering if u can help me with 2 questions please its about rms Current and emf

an ac generator has a maximum emf output of 155 v.

A. find the rms emf output(what do they mean by output)

b. fidn the rms current in the circuit when the generator is connected to a 53 resistor.

2) the largest emf that can be placed across a certain capacitor at any instant is 451 V. what is the largest rma emf that can be placed across the capacitor w/out damaging it?

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ammygurl

@ammygurl

Dazzler

+ 3

Posted: 16 years ago

#126

awww...gr8 thought....thanks guys...
now...when my school reopens..i no...i have someone to bank upon...

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Mandy Singh Walia

@Little Foot

Rocker

+ 4

Posted: 15 years ago

#127

Ohhh, till today I didn't know we have a college and uni forum! Haww! That's great. :)

I have a question regarding maths and need it....for tomorrow. (It's still not tooooo late^). I'm really having troubles.

Appraise for each real number a every real numbers x, who complete the inequation

||x-1|+x-3|<a

I hope someone kinda knows what to do and might be able to explain it to me and help me to find the right solution. I know, that there's something with 2 cases, but...

Thanks, mandy.

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akhl

@akhl

Dazzler

Posted: 15 years ago

#128

Ohhh, till today I didn't know we have a college and uni forum! Haww! That's great. :) [/quote]

Welcome to the forum. :)

It will be good if you let more and more people know about this forum so that it becomes popular.

[quote]

Appraise for each real number a every real numbers x, who complete the inequation

||x-1|+x-3|<a

[/quote]

Solution

||x-1|+x-3|<a----------(1)

The absolute value of a real number is non-negative. A non-negative number can be less than only a positive number. Therefore, a solution is possible only if a > 0

Consider the case when a > 0

Case 1: |x-1| + x-3 >= 0

(1) gives

|x-1| + x-3 < a

|x-1| + x < a+3--------(2)

Case 1i: x>=1

(2) gives

x-1+x < 3+a

2x < 4 + a

x < 2 + a/2 -------(3)

Both case 1i and case 1 are satisfied if

x-1+x-3 >= 0

2x-4 >= 0

2x >= 4

x >= 2

2 <= x ------(4)

Both (3) and (4) are satisfied if

2 <= x < 2 + a/2 --------------------(5)

For inequation (5) to be possible

2 < a/2 + 2

0 < a/2

a > 0 , which is a valid condition

Therefore, if a > 0, then 2 <= x < 2 + a/2 ----------(A)

We do not need to consider any other subcase of case 1 because 1i covers a>0, which covers all values of a for which a solution exists

Case 2: |x-1| + x-3 < 0

(1) gives

-(|x-1| + x-3) < a

|x-1| + x-3 > -a

|x-1| + x > 3 - a --------(6)

Case 2i: x>=1

(6) gives

x-1+x > 3-a

2x > 4-a

x > 2 - a/2 -------(7)

Both case 2i and case 2 are satisfied if

x-1+x-3 < 0

2x-4 < 0

2x < 4

x < 2 ------(8)

Both (7) and (8) are satisfied if

2 - a/2 < x < 2 --------------------(9)

For inequation (9) to be possible

2 - a/2 < 2

-a/2 < 0

-a < 0

a > 0, which is a valid condition

Therefore, if a > 0, then 2 - a/2 < x < 2------------(B)

We do not need to consider any other subcase of case 2 because 2i covers a>0, which covers all values of a for which a solution exists

Taking union of (A) and (B), we get

2 - a/2 < x < 2 + a/2

Ans:

If a > 0, then 2 - a/2 < x < 2 + a/2

If a <= 0, then there is no solution

Edited by akhl - 15 years ago

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Mandy Singh Walia

@Little Foot

Rocker

+ 4

Posted: 15 years ago

#129

Originally posted by: akhl

Thank you soo soo soo soo sooooo much! It's really very long, and I'm very very thankul to you that you helped me.😳😳😳

🤗*huuuugggssssss*,

mandy.
Edited by Little Foot - 15 years ago