Friction between our feet and the surface we walk on is desirable. True False
2 answers:
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Answer: 8.8 m
Explanation:
The movement of the baseball to the wall is a example of parabolic motion. While the baseball aproach the wall it is affected by gravity.
In this case, because the Initial Velocity of the ball is a vecotr, it can be defined using its two directionals compounds. One, on the Y-axis and another on the X-axis. This can be related, on how a hypotenuse is the product of two legs of the triangle. Because of this, each one of this, can be know using the following equation:
Vx = Vo * cos(∅)
Vy = Vo * sin(∅)
Where Vo is the initial velocity 27.5 m/s, and ∅ is the angle which is 20.5°. So we calculate:
Vx = 27.5m/s * cos(20.5)
Vx = 27.5m/s * 0.936
Vx = 25.74m/s
Vy = 27.5m/s * sin(20.5)
Vy = 27.5m/s * sin(20.5)
Vy = 9.62m/s
Now the movement is divided on two parts, one under the effect of gravity and another one with a constant velocity.
To know how tall does the baseball hit the wall, we need to know first how much time it takes the ball to reach the wall on the X-axis. The wall is 17.5m away, we velocity on Vx that is constant we can calculate as it follow:
Time (T) = Distance (D) / Velocity (V)
Where Distance is 17.5m and our Velocity is the Vx calculated before.
T = 17.5 m / 25.74m/s
T = 0.68s
This is the time it takes the ball to reach the wall.
Know with the time, we can calculate the how tall it got on that time with the following equation:
Where Vo is the Y-Compound of the Initial Velocity.
a is the aceleration, in this case the Gravity. Which, will be negative because is oposing the movement. Gravity is equal to 9.81
And t is the time it takes the ball to get to the wall.
This is the height that the baseball touch the wall.
Okay here initially child walks on the outer edge of the disc and then started to move inside
So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside
As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.
But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system
Here as we know that
where L = angular momentum
since here torque is ZERO
L = constant
so here we can write initial angular momentum of the system as
here we know that
= moment of inertia of merry go round
= moment of inertia of child
so here we can say
so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as
here m = mass of child
r = distance of child from center
Since child is moving inside so his distance from center is decreasing
so here moment of inertia of child is decreasing as he starts moving inside
so final angular speed of merry go round will increase as child go inside
so here as
final angular speed will be more than initial speed as child moves inside