If the forces are equal, at a distance equidistant it is not possible to act a pair on the body since both torques cancel each other. Being of the same magnitude and in the opposite direction, the sum of the torques will be zero.
1) A tiger leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from
1 answer:
The distance of tiger's leap from the base of rock is 5.58 m
It is a question of two dimensional motion
The time of motion in two dimensional motion is given by:
t=
where y is the height and g is the acceleration due to gravity
y is given to be 7.5m and let us assume g to be 9.8 m/s^2
t =
= 1.24s
Using time and speed,
We know that distance is the product of speed and time,
Distance= speed x time
speed is given to be 4.5 m/s
distance from the base of rock = 4.5 x 1.24
= 5.58m
Hence the distance of tiger's leap from the base of rock is 5.58 m
Disclaimer:
The acceleration due to gravity is assumed to be 9.8 m/s^2
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<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>
<u>Explanation:</u>
To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:
where,
are the mass, initial velocity and final velocity of ball.
are the mass, initial velocity and final velocity of bottle.
We are given:
Putting values in above equation, we get:
Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.