Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;
Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.
If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero.
Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.
The 53N of friction is (the weight) x (the coefficient of kinetic friction).
53N = (490N) x (coefficient).
Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082 .
Rounded to the nearest hundredth, that's <em>0.11 </em>. (choice 'd')
For the answer to the question above
Forecasting how a business might do in the future.
Calculating tax.
Doing basic payrolls.
Calculating Revenues.
Producing charts.
--Going past 5--
Inventory tracking
Very (VERY) basic CRM for small businesses
I hope my answer helped you.
You can only determine the speed since the only info we know is how much you ran in how long of a time.
Answer:
<em>D.) state of matter</em>
Explanation:
it can undergo chemical but not nuclear.
