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
If the force equals, for instance, 100 Newtons then 0.866 × 100 = 86.6 Newtons. This is the magnitude of the resultant force vector on the object.
Explanation:
Given that,
Mass of the cart, 
Mass of the cart 2,
Final speed of cart 2, 
Final speed of cart 1 is 0 as it comes to rest.
Let us assume that the initial velocity of the cart 2 is 0. So using the conservation of linear momentum as :

So, the initial velocity of the 1.0-kg cart is 0.9 m/s.
Answer:
1.2ms⁻²
Explanation:
F = ma
12 = 10*a
12/10 = a
6/5 ms⁻² = a or 1.2 ms⁻² = a
Answer:

Explanation:
Force 1 
Force 2 
Acceleration at stage 2 
Generally the weight of the Craft W is given as
W= upward force(thrust)
Therefore
