A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the
2 answers:
You might be interested in
Answer:
(i) 7.2 feet per minute.
(ii) No, the rate would be different.
(iii) The rate would be always positive.
(iv) the resultant change would be constant.
(v) 0 feet per min
Explanation:
Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,
By making the diagram of this situation,
Applying Pythagoras theorem,
Differentiating with respect to t ( time ),
( l = 26 feet = constant )
We have,
(i) From equation (1),
From equation (X),
(ii) From equation (X),
Thus, for different value of x the value of would be different.
(iii) Since, distance = Positive number,
So, the value of y will always a positive number.
Thus, from equation (X),
The rate would always be a positive.
(iv) The length of the ladder is constant, so, the resultant change would be constant.
i.e. x = increases ⇒ y = decreases
y = decreases ⇒ y = increases
(v) if ladder hit the ground x = 0,
So, from equation (X),
Answer:
Final velocity of the car will be -9.28 m/sec
Explanation:
We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec
Acceleration of the car in negative direction so acceleration will be
From first equation of motion we know that
v = u+at
So
So final velocity will be -9.28 m/sec