Answer:
For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.
Explanation:
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:

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,


B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

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
Work done = Force x Distance
Force = 10 lb = 44.5 N
Work Done = 44.5 N x 15 m
= 667.5 N-m
I actually know the answer to this one, you use pennies to find the atomic weight of a penny, it really doesn't have a weight. LOL
Answer:+1.25 m/s
Explanation:
Given
mass of ice skater M=70 kg
mass of ball m=10 kg
the initial velocity of the ball 
Conserving linear momentum
![M\times0+m\timesu_1=(M+m)v\quad \quad [v=\text{combined velocity of skater and ball}]](https://tex.z-dn.net/?f=M%5Ctimes0%2Bm%5Ctimesu_1%3D%28M%2Bm%29v%5Cquad%20%5Cquad%20%5Bv%3D%5Ctext%7Bcombined%20velocity%20of%20skater%20and%20ball%7D%5D)

Therefore the velocity of the person holding the ball is 1.25 m/s
This collision represents the perfectly inelastic collision where particles stick together after the collision.