• An object of mass 2.0 kg moves in uniform circular motion on a horizontal friction less table. The radius of the circle is 0.7
1 answer:
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Answer:
Explanation:
Let be the horizontal speed of the keys and D be the horizontal distance covered by the keys.
The initial velocity in the vertical direction is .
Let be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e.
will remain constant throughout the projectile motion.
Let be the total time of flight in seconds. So,
Let be the hight of the cliff.
Now, from the equation of motion
Where is the displacement is the direction of force, is the initial velocity in the direction of force, is the constant acceleration due to force and is the time of flight.
Here, , and (negative sign is for taking the sigh convention positive in the direction as shown in the figure.)
From equation (ii)
Take and put the value of
from equation (i), we have
.
Hence, the height of the cliff was .
Answer:
W=-1881.6J
Explanation:
we have that the change in the mass is
by solving the differential equation and applying the initial conditions we have
by solving for c and d
d=60
c=0.75
The work needed is
W = m(t) gh
by integrating we have
hope this helps!!