Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
Answer:
The 10 rules of badminton are as follows:
1. A game starts with a coin toss. Whoever wins the toss gets to decide whether they would serve or receive first OR what side of the court they want to be on. The side losing the toss shall then exercise the remaining choice.
2. At no time during the game should the player touch the net, with his racquet or his body.
3. The shuttlecock should not be carried on or come to rest on the racquet.
4. A player should not reach over the net to hit the shuttlecock.
5. A serve must carry cross court (diagonally) to be valid.
6. During the serve, a player should not touch any of the lines of the court, until the server strikes the shuttlecock. During the serve the shuttlecock should always be hit from below the waist.
7. A point is added to a player's score as and when he wins a rally.
8. A player wins a rally when he strikes the shuttlecock and it touches the floor of the opponent's side of the court or when the opponent commits a fault. The most common type of fault is when a player fails to hit the shuttlecock over the net or it lands outside the boundary of the court.
9. Each side can strike the shuttlecock only once before it passes over the net. Once hit, a player can't strike the shuttlecock in a new movement or shot.
10. The shuttlecock hitting the ceiling, is counted as a fault.
Explanation:
Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor, 
The static and dynamic friction coefficient is, 
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,
So, the acceleration of the car is 
. Hence, this is the required solution.
Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
= 
(
- 
) ........................................................ i
And,
v = 
v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
= ( - )
or, = (
) =() 
or, y = 3.56 cm.
