The magnetic force on a current-carrying wire due to a magnetic field is given by

where
I is the current
L the wire length
B the magnetic field strength
In our problem, L=1.0 m,

and

, so we can re-arrange the formula to find the current in the wire:
Answer:
Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.
Explanation:
#CARRYONLEARNING
<em>Convert 1nanosecond in to its SI init</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>
<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>
Answer: 321 J
Explanation:
Given
Mass of the box 
Force applied is 
Displacement of the box is 
Velocity acquired by the box is 
acceleration associated with it is 

Work done by force is 

change in kinetic energy is 

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy
![\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%2BW_f%3D%5CDelta%20K%5Cquad%20%5BW_f%3D%5Ctext%7BWork%20done%20by%20friction%7D%5D%5C%5C%5C%5C%5CRightarrow%20375%2BW_f%3D54%5C%5C%5CRightarrow%20W_f%3D-321%5C%20J)
Therefore, the magnitude of work done by friction is 