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snow_tiger [21]
3 years ago
11

Which of the following is the least important factor of a personal fitness program? A. the individual's personal conditions B. t

he availability of resources C. the level of motivation D. the time of day physical activity will be performed Please select the best answer from the choices provided. A B C D Mark this and return
Physics
1 answer:
bonufazy [111]3 years ago
8 0

Answer:

I think it's B

Explanation:

I think its trying to tell you that no matter who you are you could still do regular fitness but I don't know‍♀️

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Which shows the correct order of forces, from weakest to strongest?
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A 10-cm long solenoid has 100 turns and a radius of 5 cm. If it carries a current of 2 A, What is the magnetic field B inside th
OLga [1]

Hi there!

We can use Ampère's Law:

\oint B \cdot dl = \mu_0 i_{encl}

B = Magnetic field strength (B)
dl = differential length element (m)
μ₀ = Permeability of free space (T/Am)

Since this is a closed-loop integral, we must integrate over a closed loop. We can integrate over a rectangular-enclosed area of the rim of the solenoid - ABCD - where AD and BC are perpendicular to the solenoid.

Thus, the magnetic field is equivalent to:
\oint B \cdot dl = \int\limits^A_B {B} \, dl  + \int\limits^B_C {B} \, dl   + \int\limits^C_D {B} \, dl   + \int\limits^D_A {B} \, dl

Since AD and BC are perpendicular, and since:
\oint B \cdot dl = B \cdot L = BLcos\phi

BLcos(90) = 0

If perpendicular to the field, the equation equals 0.

Additionally, since AB is outside of the solenoid, there is no magnetic field present, so B = 0. The only integral we integrate now is:
\oint B \cdot dl = \int\limits^C_D {B} \, dl

Which is horizontal and inside the solenoid. Let the distance between C and D be 'L', and the enclosed current is equivalent to the number of loops multiplied by the current:

B L = \mu_0 Ni

N = # of loops per length multiplied by the length, so:
BL = \mu_0 nL i \\\\B = \mu_0ni

Plug in the given values and solve. Remember to convert # of loops to # of loops per unit length.

B = \mu_0 (100/0.1)(2) = (4\pi *10^{-7})(1000)(2) = \boxed{0.00251 T}

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