21. light changes its direction when travelling through a new medium because denser mediums have a higher angle of refraction.
(A) We can solve the problem by using Ohm's law, which states:

where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know

and

, therefore we can rearrange Ohm's law to find the current through the device:

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:

where

is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
Answer:
24cm/s
Explanation:
A=L*w
A'=L'*w'
L=13
w=5
L'=4
w'=6
A=?
A'=?
A=L*w
A=13*5
A=65
A'=L'*w'
A'=4*6
A'=24
*the given lengths are just to throw you off*
1st <span>the total </span>energy<span> of an </span>isolated system<span> is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. ▲U=Q-W
</span><span>
2nd the total </span>entropy<span> can never decrease over time for an </span>isolated system, that is, a system in which neither energy nor matter can enter nor leave.
DS (Greater than or equal to) 0