A bowling ball has a mass of 10 kilograms. A tennis ball has a mass of 0.08 kilogram. How much inertia does the bowling ball hav
1 answer:
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<u>Answer:</u>
a) Speed of mailbag after 3 seconds = 32.4 m/s
b) Package is 44.1 meter below helicopter
c) If the helicopter was rising steadily at 3.00 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Package is 44.1 meter below helicopter
<u>Explanation:</u>
a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Initial velocity = 3 m/s, acceleration = 9.8 and time = 3 seconds.
v = 3+9.8*3 = 32.4 m/s
Speed of mailbag after 3 seconds = 32.4 m/s
b) We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 53.1 meter.
So package is (53.1-9)meter below helicopter = 44.1 m
c) Initial velocity = -3 m/s, acceleration = 9.8 and time = 3 seconds.
v = -3+9.8*3 = 26.4 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 35.1 meter.
So package is (35.1+9)meter below helicopter = 44.1 m
a) Potential energy: 147 m [J]
The gravitational potential energy of an object is given by
where
m is its mass
is the acceleration of gravity
h is the height of the object above the ground
In this problem,
h = 15 m
We call 'm' the mass of the ball, since we don't know it
So, the potential energy of the ball at the top of the hill is
(J)
b) Velocity of the ball at the bottom of the hill: 17.1 m/s
According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:
where
v is the final velocity of the ball
We know from part a) that
U = 147 m
Substituting into the equation above,
And re-arranging for v, we find the velocity: