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QveST [7]
3 years ago
9

A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left the person's hand at

-5.0 m/sec, what was its velocity on impact?
Physics
1 answer:
GrogVix [38]3 years ago
5 0
The acceleration of gravity on Earth is 9.8 meters per second every second.  So
in 1.1 second of falling, an object gains (9.8 x 1.1) = 10.78 meters per second of
downward speed.

The balloon was already falling at 5 m/sec when the time began, and it gained
another 10.78 m/sec on the way.  So at the end of 1.1 second, its downward
speed was (5.0 + 10.78) = 15.78 m/sec.

Its velocity was 15.78 m/sec downward. 
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Calculate the angular momentum, in kg · m2/s, of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.310 kg ·
steposvetlana [31]

Answer:

a) 11.7 kg. m2/s b) 0.76 Kg. m2 c) -0.33 N.m

Explanation:

a)

  • Assuming no external torques act on the skater, total angular momentum must be conserved:

        L1 = L2

        As the angular momentum can be calculated as the

        product of the moment of inertia times the angular velocity,

we can write:

       I1*ω1 = I2*ω2  

       The initial angular momentum can be written as follows:

       I1*ω1= 0.31 kg.m2 * 6.0 rev/sec  

       As we need to express the angular momentum in kg.m2/s, we need to convert the angular velocity units, from rev/sec to rad/sec, as follows:

       ω1= 6.0 rev/sec (2π rad/ rev) = 12 π rad/sec

       I1*ω1= 0.31 kg.m2 * 12 π rad/sec = 11.7 kg. m2/s

b)

  • As the final angular momentum must be the same, and we know the value of the final angular velocity, we can replace by the values in L2, and solve for I2, as follows:

        I2 = I1*( ω1 / ω2) = 0.31 kg. m2 . 6.0/2.45 = 0.76 kg.m2

c)

  • If an external torque is present, we can write the following equation, that relates the external torque with the rotational inertia and the angular acceleration, as follows:

        Τ= I *γ (1)

        Where γ, is the angular acceleration.

        By definition, γ is the rate of change of the angular velocity,

        so if we have the values of  the initial and final angular

        velocities, and the time passed, we can express γ as

follows:

       γ= (ω2 – ω1) / t

       In order to express γ in rad/sec2, we need to convert the

angular  velocities (given in rev/sec), to rad/sec, as follows:

       ω1= 6.0 rev/sec (2π rad/ rev) = 12 π rad/sec

       ω2= 3.0 rev/sec (2π rad/ rev) = 6 π rad/sec

       Solving for γ:

       γ = -6 π / 18. 0 rad/sec2 = -1.05 rad/sec2

       Replacing in (1), we have:

       τ= 0.31 kg. m2.*(-1.05 rad/sec2) = -0.33 N.m

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