Answer:
C) 10 cm
Explanation:
We shall first calculate the focal length of the mirror.
object distance u = 20 cm ( negative )
Image distance v = 12 cm ( negative )
using mirror formula
1/ v + 1 /u = 1/ f
-1/12 - 1/ 20 =1/ f
f =- 7.5 cm
In the second case
u = - 30 cm
f = - 7.5 cm
i / v + 1 / u = 1 / f
1 / v - 1/30 = - 1/ 7.5
v = -10 cm
Answer is C that is 10 cm
Answer:
Part a)

Part b)

Part c)

Explanation:
As we know that car start from rest and reach to final speed of 87 mph
so we have

now we have
Part a)
acceleration is rate of change in velocity



Part b)
distance moved by car with uniform acceleration is given as



Part c)
As we know that the car start from rest
so final speed after t = 10 s



There are 5 atoms 1=LI 1=N 3=O
Using the equation v(avg)=distance/time
and the equation v=v(original)+a(t)
solve for acceleration
2600=0+a(12)
a=216.66666 m/s^2
Then, you use the equation
v^2=v(original)+2a*(change in x)
2600^2=2(216.666666)*change in x
6760000/2/216.666666 = 15600 meters which is the length of the race
Then using v(avg)=x/t
15600/12= 1300 m/s
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball