Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
x = ?
(7-0)/3.6 = t
t = 1.94 s
Explanation:
It is given that, the position of a particle as as function of time t is given by :
Let v is the velocity of the particle. Velocity of an object is given by :
![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)
So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :
![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)
At t = 0, 
Hence, this is the required solution.
The correct answer is "C". 'Old theories are adjusted to incorporate all old new information.' This makes the most sense, regarded the old and new information should be taken into consideration.
I hope this helped you!
Brainliest answer is always appreciated!
