Answer:
true
Explanation:
I did this unit for science
In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, to increase the potential energy, the iron will move towards the electromagnet.
<h3>Potential energy of a system of magnetic dipole</h3>
The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

where;
is the dipole moment- B is the magnetic field


Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;
- In step 1, to increase the potential energy, the iron will move towards the electromagnet.
- In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.
Learn more about potential energy in magnetic field here: brainly.com/question/14383738
Answer:
X-rays go all the way through the body, but ultraviolet rays do not.
Explanation:
An x-ray will show inside the body, but uv light isn't strong enough to go all the way through the body.
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux

Put the value into the formula


We need to calculate the induced emf
Using formula of induced emf

Put the value into the formula

Put the value of emf from ohm's law





We know that,


We need to calculate the voltage across the capacitor
Using formula of charge


Put the value into the formula


Hence, The voltage across the capacitor is 1.57 V.