Answer:
b is the answer
Explanation:
tq friend b is the answer
Answer:
Explanation:
a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .
First condition of equilibrium is that net force on the body should be zero.
or F net = 0
A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0
So first condition of equilibrium is not satisfied.
Second condition is that , net torque acting on the body must be zero.
In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie towards the line joining the body under rotation with the axis .
F is along r
torque = r x F
= r F sinθ
θ = 0 degree
torque = 0
Hence 2nd condition is fulfilled.
Answer:
Explanation:
In this question we have given
we have to find
We know that
optical path difference for bright fringe is given as
Here,
n is order of fringe
and optical path difference for dark fringe is given as
since the light with wavelength produces its third-order bright fringe at the same place where the light with wavelength produces its fourth dark fringe
it means
optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe
Therefore,
...............(1)
Put value of in equation (1)
Answer:
Masses and distance between them
Explanation:
The gravitational force between two objects can be calculated using Newton's Gravitational Law.
However, using logic, we can already dictate what the answer will be, for example. We know that the bigger an object is, the stronger its gravity is. This can be seen with how the moon is much smaller, and also has much less gravity.
Also, the distance between two objects also influences the gravity. This can be seen the further an object gets from Earth, the less of a pull the gravitational field has on it. Another example is that Pluto (being very far from the sun) has less of a gravitational effect from the sun, in comparison to Mercury (the closest plant to the sun).
Answer:
total work is = 52450 J
Explanation:
given data
mass = 5000-lb
density = 10 lb/ft
height = 50 ft
solution
as we will treat here cable and ball are separate
and
here work need to lift cable is
w = (10Δy )(9.8 y ) j
and
now summing all segment of cable
so passing limit Δy to 0
so total work need
=
=
= 2450J
so lifting 5000 lb wrcking 50 m required additional 5000 + 2450
so total work is = 52450 J