It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
The maximum torque is
Explanation:
From the question we are told that
The length of the wire is 
The current flowing through the wire is 
The magnetic field is 
The maximum torque is mathematically evaluated as
Where
is the magnetic dipole moment which is mathematically represented as

Where
is the number of turns which from the question is 1
substituting values


Now
The smallest unit of an element that has all the properties of the element is an atom. This has all the properties of an atom, including the particles.
Answer:
A. Rocket A will travel farther horizontally than rocket B.
Explanation:
This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.
Answer:
0.0389 cm
Explanation:
The current density in a conductive wire is given by

where
I is the current
A is the cross-sectional area of the wire
In this problem, we know that:
- The fuse melts when the current density reaches a value of

- The maximum limit of the current in the wire must be
I = 0.62 A
Therefore, we can find the cross-sectional area that the wire should have:

We know that the cross-sectional area can be written as

where d is the diameter of the wire.
Re-arranging the equation, we find the diameter of the wire:
