The potential difference between points a and b is zero.
<h3>Total emf of the series circuit</h3>
The total emf in the circuit is the sum of all the emf in the circuit.
emf(total) = 1.5 + 1.5 = 3.0 V
<h3>Potential difference</h3>
The potential difference between two points, a and b is calculated as follows;
V(ab) = Va - Vb
V(ab) = 1.5 - 1.5
V(ab) = 0
Thus, the potential difference between points a and b is zero.
Learn more about potential difference here: brainly.com/question/3406867
I can't eliminate answers. Some of them are just wrong. A is incorrect. There is no such thing as a 1 pole magnet.
I wouldn't use B. If it is just a bar it is not a magnet.
C is the traditional answer
D is a space filler. It is just there to occupy a letter.
Answer:
Kidneys filter our blood,
Explanation:
Hope this helped :)
That was a lucky pick.
Twice each each lunar month, all year long, whenever the Moon,
Earth and Sun are aligned, the gravitational pull of the sun adds
to that of the moon causing maximum tides.
This is the setup at both New Moon and Full Moon. It doesn't matter
whether the Sun and Moon are both on the same side of the Earth,
or one on each side. As long as all three bodies are lined up, we
get the biggest tides.
These are called "spring tides", when there is the greatest difference
between high and low tide.
At First Quarter and Third Quarter, when the sun, Earth, and Moon form a
right angle, there is the least difference between high and low tide. Then
they're called "neap tides".
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = 
= - 
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object
