A golf ball is sitting on a tee. The ball is struck with a golf club and flies
2 answers:
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Explanation:
- Newton's first law of motion:
"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force
In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.
- Newton's second law of motion:
"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:
where F is the net force, m is the mass of the object, and a its acceleration"
In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.
"Newton's third law of motion:
"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.
1 kilometre=1000 metre
1 hour = 3600 second
The initial velocity of car A is 35.0 km/hr i.e
= 9.72 m/s
The initial velocity of car B is 45 km/hr =12.5 m/s
The initial velocity of car C is 32 km/hr = 8.89 m/s
The initial velocity of car D is 110 km/hr=30.56 m/s
The acceleration of car A is given as
The time taken by car A = 15 min.
From equation of kinematics we know that-
v= u+at [Here v is the final velocity and a is the acceleration and t is the time]
Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s
=11.456111106 m/s
The acceleration of B is given as
The time taken by car B =20 min
The final velocity of B is -
v= u+at
= u-at [Here a is negative due to deceleration]
=12.5 m/s +[0.0011574074074×20×60]
=13.8888888.....
=13.9
The acceleration of C is given as
The time taken by car C =30 min
The final velocity of C is-
v = u+at
=8.89 m/s+[0.003086419753×30×60] m/s
=14.4455555555..m/s
=14.45 m/s
The car C is decelerating.The deceleration is given as-
The time taken by car D= 45 min.
The final velocity of the car D is-
v =u+at
=30.56 -[0.00462962962962×45×60]m/s
=18.06 m/s
Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.