Answer:
42.0×10² second²
Explanation:
Here, time is given in milisecond
(64800 ms)²
= 4199040000 ms²
The SI unit is seconds
1 second = 1000 milisecond
![1\ milisecond=\frac{1}{1000}\ second](https://tex.z-dn.net/?f=1%5C%20milisecond%3D%5Cfrac%7B1%7D%7B1000%7D%5C%20second)
![\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%201%5C%20milisecond%5E2%3D%5Cleft%28%5Cfrac%7B1%7D%7B1000%7D%5Cright%29%5E2%5C%20second%5E2)
![4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2](https://tex.z-dn.net/?f=4199040000%5C%20ms%5E2%3D4199040000%5Ctimes%20%5Cleft%28%5Cfrac%7B1%7D%7B1000%7D%5Cright%29%5E2%5C%20second%5E2%3D4199.04%5C%20second%5E2)
42.0×10² second²
Answer:
i think it's 2km pm
Explanation:
2km x 30 60.. start was 30, and now your at 90.. we had to determine how much time it took.. so 2 is the average.. or atleast per minute and sorry it i still didnt answer ur question lol im just trynna help
To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.
The diple moment associated with an iron bar is given by,
![\mu = \alpha *N](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Calpha%20%2AN)
Where,
Dipole momento associated with an Atom
N = Number of atoms
y previously given in the problem and its value is 2.8*10^{-23}J/T
![L = 5.8cm = 5.8*10^{-2}m](https://tex.z-dn.net/?f=L%20%3D%205.8cm%20%3D%205.8%2A10%5E%7B-2%7Dm)
![A = 1.5cm^2 = 1.5*10^{-4}m^2](https://tex.z-dn.net/?f=A%20%3D%201.5cm%5E2%20%3D%201.5%2A10%5E%7B-4%7Dm%5E2)
The number of the atoms N, can be calculated as,
![N = \frac{\rho AL}{M_{mass}}*A_n](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B%5Crho%20AL%7D%7BM_%7Bmass%7D%7D%2AA_n)
Where
Density
Molar Mass
A = Area
L = Length
Avogadro number
![N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)](https://tex.z-dn.net/?f=N%20%3D%20%5Cfrac%7B%287.9g%2Fcm%5E3%29%281.5cm%29%285.8cm%5E2%29%7D%7B55.9g%2Fmol%7D%286.022%2A10%5E%7B23%7Datoms%2Fmol%29)
![N = 7.4041*10^{23}atoms](https://tex.z-dn.net/?f=N%20%3D%207.4041%2A10%5E%7B23%7Datoms)
Then applying the equation about the dipole moment associated with an iron bar we have,
![\mu = \alpha *N](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Calpha%20%2AN)
![\mu = (2.8*10^{-23})*(7.4041*10^{23})](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%282.8%2A10%5E%7B-23%7D%29%2A%287.4041%2A10%5E%7B23%7D%29)
![\mu = 20.72Am^2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2020.72Am%5E2)
PART B) With the dipole moment we can now calculate the Torque in the system, which is
![\tau = \mu B sin(90)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cmu%20B%20sin%2890%29)
![\tau = (20.72)(2.2)](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%2820.72%29%282.2%29)
![\tau = 45.584N.m](https://tex.z-dn.net/?f=%5Ctau%20%3D%2045.584N.m)
<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>
Answer:
The answer would be Igneous rock
Explanation: