The body will take 20 seconds to cover a distance of 1000 m i.e. 1 km
It is D, the distance between each object squared. We get this from solving for the force of gravity, which is
F = G((m1*m2)/(r^2)).
Here, you can see that the masses of the two objects are divided by their total distance apart from each other (r) squared.
Answer:
Explanation:
Given the forces:
F1 = 8.92 i + 17.37 j
F2 = 8.31 i - 10.97 j
If a third vector is added to them such that they add up to to the null vector as F1 + F2 + F3 = 0
then to get F3:
F3 = -F2-F1
F3 = -(8.31 i - 10.97 j)-(8.92 i + 17.37 j
)
F3 = -8.31 i + 10.97 j-8.92 i - 17.37 j
F3 = -8.31i-8.92i+10.97j-17.37j
F3 = -17.23i-6.4j
from the vector:
x = -17.23 and y = -6.4
angle of the third vector with respect to the +x-axis is expressed as:
Hence the angle the vector makes with the x axis will be 
Answer:
1. The image is real
2. 5.85
3. h' = 3.05 mm
4. The image is upright
Explanation:
1. Start with the first lens and apply 1/f = 1/p + 1/q
1/5.01 = 1/13.7 + 1/q
q = 7.90 cm
Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,
1/25.9 = 1/54.6 + 1/q
q = 49.3 cm behind the second lens
Using that information, since q is positive, the image is real
2. Also, using that information, you have the second answer, which is 49.3 cm
The height can be found from the two magnifications.
m = -q/p
m1 = -7.9/13.7 = -.577
m2 = -49.3/54.6 = -.903
Net m = (-.577)(-.903) = .521
Then, m = h'/h
.521 = h'/5.85
3. h' = 3.05 mm
4. For the fourth answer, since the overall magnification is positive, the final image is upright
