Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as





FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so




Change in entropy 
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Mechanical digestion is chewing, and chemical digestion is the saliva in your mouth breaking down food.