All Answers 355911 Answers

Reposted here:

This is a repost of this question:

So the peltier cooler will be fine at 9 volts but the fan will fry very quickly? In that case would i need to include a ic 7805 right before it reaches the fan and then connect it back to the main circut. But when the 5 volt wire gets re-connects with the 9 volt wire will nothing happen? I also dont want to burn the battery fast.

It is fine if the peltier works at the lower standards in terms or temperature. It is not cooling something that will become too warm.

I do not think you will have good luck with that.    Running the peltier at lower voltage than spec means you will not get the spec'd cooling from it......this may be ok for your application....but not good if you need the specified amount of cooling. Shortened component life would be expected if inadequate cooling is supplied.   Running the 5 v fan with the 9 volt battery will likely work for a short while (perhaps very short while) until something burns.     Get a 12 v battery and a 12 volt fan or run the fan with a resistor in-line with it to limit current .

My 2 cents.     You can always experiment ...if you can afford some fried components.    Others may be much more knowledgeable on this forum.

 LOL 

XD

Understood. I'll ask the professor about it and reply here again when I find more info

Regarding the question itself, that is all she gave me. You can probably see that "the hanging cables make 60° with the sliding cable" isn't very specific at all. Otherwise, I'm not really sure at all how to get the correct degree values here. Is using that X-axis as a reference incorrect?

We answer physics probems all the time (when we can)....post on!  

If hanging cable makes 60 degrees with sliding cable.....how can sliding cable be at 30 degrees ?      That would make hanging cable at -90 degrees....which it is not.     Need actual sliding cable degrees..........or the degrees between the hanging cables.   Do you have the correct numbers?

An even perfect square is produced by an even number. The smallest, being 2, which 2^2 = 4.

Then, the largest, being 30, which is 30^2=900.

Counting all of the even numbers from 2-30 yields your answer is A.15. 

The size of your graph paper.           JK  

Answer:

speed: 5 m/s
height: 9.99 m 

Explanation:

KE=12mv2
kinetic energy = 12 * mass * velocity squared

12.5J=12⋅1kg⋅v2
1⋅v2=25J
v2=25J
v=±5J

since speed does not specify direction as velocity does, we can take the value of 5m/s for speed.

GPE=mgh
gravitational potential energy = mass * gravitational field strength * height

assuming the airplane is flying above Earth, the gravitational field strength is 9.81N/kg.

98J=1kg⋅9.81J⋅h

98J=9.81h

9.81h=98

h=989.81=9.99

h=9.99m 

3/5 = 12/x, x=20.

read this: https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_4

First, multiply by the conjugate of 1-x, which is \(\frac{1-x^2}{1+x}\). 

Then, simplifying things, we get \(\lim\limits_{x\rightarrow1}(-\frac{x \ln x}{x-1})\).

Which then turns into this: \(-\lim _{x\to \:1}\left(\frac{\frac{1}{-1+x}\ln \left(x\right)}{\frac{1}{x}}\right) \\ \frac{\lim _{x\to \:1}\left(\frac{1}{-1+x}\ln \left(x\right)\right)}{\lim _{x\to \:1}\left(\frac{1}{x}\right)} \)

Using L'Hopital's rule and simplifying, we get that this equation is equal to 1. 

We also need to find the value of \(\lim _{x\to \:1}\left(\frac{1}{x}\right)\) which is easily equal to 1.

Therefore your answer is \(-\frac{1}{1} = \boxed{-1}\).

Pretty rampant parenthetical lack on this site ! 

GPE = mgh

39200  = 200 kg  * 9.81 m/sec^2  * h        Solve for 'h'

39200/(200*9.81) = h meters       

m must be four, since 4^2 = 16. Then, 2mn = degree 1 term, which means mn = 18.

50 times 25 - 50 - 50 times 5 - 50 times 10 = 50 times 9.

A is your answer.

I wouldn't let it be \(e^{x-1}\) without a thorough scolding about using parentheses!

Give an example of what you mean by "at most two different digits".

Jack rolls 5 fair six-sided dice. What is the probability that at least two dice show the same number?

1st die - any number, or.................................................................. = 6/6
2nd die-probability that it is different from the first......................     =5/6
3rd die-................................................................................1st two =4/6
4th die-...............................................................................1st three =3/6
5th die-................................................................................1st four =2/6
So, the probability that they are all different from each other =6/6 x 5/6 x 4/6 x 3/6 x 2/6 =6! / 6^5 =0.09259.
And the probability that at least 2 of them show the same number =1 - 0.09259 =90.74%

It only has the calc on the home page.