Mechanical wave shows dual nature
Answer:

Explanation:
The torque of a force is given by:

where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation of the system
is the angle between the direction of the force and d
In this problem, we have:
, the force
, the distance of application of the force from the centre (0,0)
, the angle between the direction of the force and a
Therefore, the torque is

Answer:
a) Acceleration of the car is given as

b) Acceleration of the truck is given as

Explanation:
As we know that there is no external force in the direction of motion of truck and car
So here we can say that the momentum of the system before and after collision must be conserved
So here we will have

now we have


a) For acceleration of car we know that it is rate of change in velocity of car
so we have



b) For acceleration of truck we will find the rate of change in velocity of the truck
so we have



Answer:
The answer is 24cm
Explanation:
This problem bothers on the curved mirrors, a concave type
Given data
Object height h= 5cm
Object distance = 12cm
Focal length f=24cm
Let the image distance be v=?
Applying the formula we have
1/v +1/u= 1/f
Substituting our given data
1/v+1/12=1/24
1/v=1/24-1/12
1/v=1-2/24
1/v=-1/24
v= - 24cm
This implies that the image is on the same side as the object and it is real