The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:

Where,
Final and initial velocity
a = Acceleration
x = Displacement
For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore


Our values are given as


Replacing we have that,



Therefore the speed with which the liquid sulfur left the volcano is 529.15m/s
Answer:
Approximately
(assuming that the melting point of ice is
.)
Explanation:
Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

The energy required comes in three parts:
- Energy required to raise the temperature of that
of ice from
to
(the melting point of ice.) - Energy required to turn
of ice into water while temperature stayed constant. - Energy required to raise the temperature of that newly-formed
of water from
to
.
The following equation gives the amount of energy
required to raise the temperature of a sample of mass
and specific heat capacity
by
:
,
where
is the specific heat capacity of the material,
is the mass of the sample, and
is the change in the temperature of this sample.
For the first part of energy input,
whereas
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
Similarly, for the third part of energy input,
whereas
. Calculate the change in the temperature:
.
Calculate the energy required to achieve that temperature change:
.
The second part of energy input requires a different equation. The energy
required to melt a sample of mass
and latent heat of fusion
is:
.
Apply this equation to find the size of the second part of energy input:
.
Find the sum of these three parts of energy:
.
It takes more work to use a meat grinder
Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as

for uniform cylinder we will have

so we have


now we have




now we have


Part b)
speed of the bucket can be found using kinematics
so we have



Part c)
now in order to find the time of fall we can use another equation



Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance


