Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.
If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2
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
Answer:twice of initial value
Explanation:
Given
spring compresses
distance for some initial speed
Suppose v is the initial speed and k be the spring constant
Applying conservation of energy
kinetic energy converted into spring Elastic potential energy

When speed doubles

divide 1 and 2


Therefore spring compresses twice the initial value