Crushing increases the ice's surface area. Additionally, cooling from ice happens as a result of convection, which is proportional to area. Thus, breaking up the ice expedites the cooling process.
<h3>What is cooling process?</h3>
When precise and consistent temperature control inside a process is necessary, this sort of cooling is used. Because they can offer cooling capacity independent of variations in the ambient temperature, heat load, and flow needs of the application, chillers are frequently employed to remove heat from a process.
It's important to properly regulate the heat load in many industrial operations. Different components of the whole application might need to be cooled.
Major locations for cooling include:
Direct product cooling
During the molding process, plastic
milling metal products
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The resistance of a conductor is given by

where L is the length of the wire,

the resistivity of the material and A the cross-sectional area.
We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length:

, then the new resistance is also 4 times the original value:
Answer:
Acceleration due to gravity will be 
Explanation:
We have given length of pendulum l = 55 cm = 0.55 m
It is given that pendulum completed 100 swings in 145 sec
So time taken by pendulum for 1 swing 
We have to find the acceleration due to gravity at that point
We know that time period of pendulum;um is given by

So 

Squaring both side


So acceleration due to gravity will be 
Answer:
179.47m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute
7750(179)+72(230) = (7750+72)v
1,387,250+16560 = 7822v
1,403,810 = 7822v
v = 1,403,810/7822
v= 179.47m/s
Hence the final velocity of the probe is 179.47m/s
We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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