Aayushka karki

Though there is no gain in mechanical advantage by using a single fixed pulley it is used in practice, it is because of the following reasons:

• It is simpler to learn and use.
• The entire weight of the person is used in doing work.
• It is easier to construct.
  
  

Mechanical advantage (MA) is theratio of Load to Effort. Both load and effort are forces, i.e., MA is the ratio of two same physical quantities. So, it has no unit.

Solution:

Initial velocity of stone (u) = 72 km/hr = 72 * 1000/3600 = 20 m/s

The stone is acted by constant downward acceleration equal to the acceleration due to gravity of earth, i.e, a = -g = - 10 m/s²

At maximum height, the velocity of the stone becomes 0 for a very short time so, final velocity of stone (v) = 0 m/s

To find: maximum height (H) = ?

Using equations of motion,

v² = u² + 2aH

or, 0² = (20)² + 2(-10)H

or, 0 = 400 - 20H

or, 20H = 400

or, H = 400/20

So, H = 20 m

Hence, the maximum height...

Forces are said to be balanced when there is no net push or pull acting on the body experiencing the force. Example: balanced condition of tug of war, etc.

Forces are said to be unbalanced when there is some push or pull acting on the body experiencing the force. Example: when a team pulls another team in a tug of war, etc.

The fan tends to continue its circular motion even after it is switched off due to the inertia of motion. The fan rotates with a decrease in speed (this is because of air friction) and stops after a few minutes.

Given

initial velocity ( u ) = 72 km/hr = 72 x 1000 / 3600

= 20 m/s

acceleration due to gravity (g) = 10 m/s²

also, final velocity(v) = 0 m/s

for maximum height,

v² = u² - 2gH

or, 0 = 20² - 2 x 20 H

thus, H = 400/ 40

maximum height (H) = 10 m

again , for time to return to initial position.

total displacement(s) = 0

so,

s= ut - 1/2 gt²

or, ut = gt²

thus, T = u/g

= 20/10

total time taken = 2 sec

Solution:

Initial velocity of car (u) = 45 km/hr = 45 * 1000/3600 = 12.5 m/s

Time taken (t) = 3 s

Since the car stops finally, final velocity of car (v) = 0 m/s

Now,

Retardation (-a) = (u - v)/t

= (12.5 - 0)/3

= 4.17 m/s²

Hence, the magnitude of retardation of the car is 4.17 m/s².

Let the car move a distance 's' before coming into rest than from equations of motion, we have,

s = ut - 1/2 (a)t² [acceleration is negative due to retardation]

= (12.5)3 - 1/2 (4.17) (3)²

= 18.375 m

Hence, the car travelled...