Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is

Equating initial and the final momenta we get

Now since the surface is frictionless thus the energy is also conserved thus

Similarly the final energy becomes
\
Equating initial and final energies we get

Solving i and ii we get

Using this in equation i we get
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.
Hi there!
In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.
The net force is equivalent to the centripetal force, so:
∑F = T
mv²/r = T
Solve for v:
v = √rT/m
v = 13.96 m/s
Answer:
option A
Explanation:
given,
height of the drop of stone = 9.44 m
speed of the stone = ?
As the stone is dropped the energy of the stone will be conserved.
using conservation of energy.
Potential energy = Kinetic energy

v = 13.60 m/s
Hence, the correct answer is option A
Answer:
The time for the cake to cool off to room temperature is
approximately 30 minutes.
Let
=
F be the temperature and T that of the body
Explanation:
Our Tm = 70, the initial-value problem is
= <em>k</em>(T − 70), T(0) = 300
Solving the equation, we get
= <em>kdt</em>
In [T-70]= <em>kt </em>+
T = 70 +

Finding he value for
using the initial value of T (0)= 300, therefore we get:
300=70+
= 230 therefore
T= 70+ 230 
Finding the value for <em>k </em>using T (3) = 200, therefore we get
T (3) = 200
= 
<em>K </em>=
in 
= -0.19018
Therefore
T(t) = 70+230