Answer:
A) q = -8.488 cm
, B) m = 0.29
Explanation:
A) For this exercise in geometric optics, we will use the equation of the constructor
where p and q are the distance to the object and image, respectively and f is the focal length
in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

we calculate

= - 0.1178
q = -8.488 cm
the negative sign indicates that the image is virtual
B) the magnification is given

we substitute
m =
m = 0.29
the positive sign indicates that the image is right
The direction of work.........
Answer:
78.4 m
Explanation:
Using newton's equation of motion,
S = ut + 1/2gt²......................... Equation 1
Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.
Note: Taking upward to be negative, and down ward positive
Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²
Substitute into equation 1
S = 49(2) - 1/2(9.8)(2)²
S = 98 - 19.6
S = 78.4 m
Hence the height of the ball two seconds later = 78.4 m
Explanation:
According to Newton's First Law of motion, if a box is pushed with no external resistance, the box will keep on moving due to the absence of external force. It might gets stopped due to frictional force that is acting between the surface and the ball. The first law of motion is also known as law of inertia. the magnitude of force acting on the object is given by second law of motion.