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3 years ago

I need help with air resistance...

1 answer:

3 0

Mar 4, 2004 #1 FabioTTT the formula: R = (.5)DpAv^2 is used to determine how much air resistance (resistive force) is being placed on the object. R is the Resistive force. D is some dimensionless empirical quantity called the drag coefficient p is the density of air A is the cross-sectional area of the object (surface area) v is velocity My question is, is the density of air some constant that should be already given to me? also, how would i go about finding the drag coefficient? I'm asking this because in physics we're doing a lab in wich we drop a cofee filter... and we record its time to reach a certain height (which im guessing is to be able to calulate for the terminal velocity). So in other words, I have all of these variables except for R, D, and p. Can anyone help me out? Phys.org - latest science and technology news stories on Phys.org

Reference https://www.physicsforums.com/threads/need-help-with-air-resistance.15678/

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3 years ago

A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

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