<span>What is the main fuel consumed in the core of a red giant?
The </span><span>main fuel consumed in the core of a red giant is He or helium. The answer is letter D.</span>
I think you forgot to give the options along with your question. I am answering the question based on my knowledge and research. <span>A business that sells products to teens would most likely create a website with a title ending in .com. I hope that this is the answer that has actually come to your great help.</span>
Weight = (mass) x (gravity)
Weight = (7.0 kg) x (gravity)
On Earth, where (gravity) is roughly 10 N/kg . . .
Weight = (7.0 kg) x (roughly 10 N/kg)
Weight = roughly 70 Newtons
That's <em>B </em>on Earth.
It would be some other number on other bodies.
The answer is Column, however if you want to know the scientific word for it, it is also known as a group
(extra fact, a row is called a period)
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
![\frac{1}{q}=\frac{1}{f}-\frac{1}{p}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7Bf%7D-%5Cfrac%7B1%7D%7Bp%7D)
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
![\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7B-10%20cm%7D-%5Cfrac%7B1%7D%7B15%20cm%7D%3D-0.167%20cm%5E%7B-1%7D)
![q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B1%7D%7B-0.167%20cm%5E%7B-1%7D%7D%3D-6.0%20cm)
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
![h_i = - h_o \frac{q}{p}](https://tex.z-dn.net/?f=h_i%20%3D%20-%20h_o%20%5Cfrac%7Bq%7D%7Bp%7D)
where
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual