Answer:
0Nm, no work is done.
Explanation:
Work done is defined as the Force per distance meaning force times the distance moved in the direction of the force.
Now the body of mass 50g has a centripetal force acting on it directed towards the centre. Now in actuality the body stays along the circle it doesn't really move to the centre of the circle.
Hence the force doesn't move a distance, and so from the definition of work done;
F×d ; d =0
Hence work done = mv2/r × 0= 0Nm
Formula:
F = ma
F: force (N) m: mass (kg) a: acceleration (m/s^2)
Solution:
F = ma
F = 20 × 10
= 200N
The speed of the rock at 20 m is 34.3 m/s
Explanation:
We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:
where
:
is the initial potential energy
is the initial kinetic energy
is the final potential energy
is the final kinetic energy
The equation can also be rewritten as follows:
where:
m = 100 kg is the mass of the rock
is the acceleration of gravity
is the initial height
u = 0 is the initial speed (the rock starts at rest)
is the final height of the rock
v is the final speed when h = 20 m
And solving for v, we find:

Learn more about kinetic energy and potential energy here:
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The energy stored in motion is called kinetic energy.
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:

Where,
F =View Factor
A = Cross sectional Area
Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m

The view factor between two coaxial parallel disks would be


Then the view factor between base to top surface of the cylinder becomes
. From the summation rule


Then the net rate of radiation heat transfer from the disks to the environment is calculated as





Therefore the rate heat radiation is 780.76W