Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2. We want
to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance. Which kinematic formula would be most useful to solve for the target unknown?
1 answer:
Explanation:
It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².
The second equation of kinematics gives the relationship between the height reached and time taken by it.
Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.
We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :
t is time taken by the ball to hit the ground
is initial speed of the ball
So, the correct option is (A).
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Answer:
<em>a) 6738.27 J</em>
<em>b) 61.908 J</em>
<em>c) </em>
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Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
=
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
=
= 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = = 6.655 x
= <em>6738.27 J</em>
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b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
=
=
= 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
where the subscripts 1 and 2 indicates the values first and second flywheels
= (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = = 6.655 x
= <em>61.908 J</em>
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c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy =
where m is the mass of the car
is the velocity of the car
Equating the energy
2246.09 =
making m the subject of the formula
mass of the car m =