Answer:
The entropy change of the sample of water = 6.059 x 10³ J/K.mol
Explanation:
Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol
Mathematically, entropy is expressed as
ΔS = ΔH/T....................... Equation 1
Where ΔH = heat absorbed or evolved, T = absolute temperature.
<em>Given: If 1 mole of water = 0.0018 kg,</em>
<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>
<em>T = 100 °C = (100+273) K = 373 K.</em>
<em>Substituting these values into equation 1,</em>
<em>ΔS =2.26x 10⁶/373</em>
ΔS = 6.059 x 10³ J/K.mol
Therefore the entropy change of the sample of water = 6.059 x 10³ J/K.mol
<span>So we want to know what happens to the momentum of the ball that rolls down hill and hits a box. So we need to use the law of conservation of momentum which states that the momentum must be conserved. It cant be transformed into inertia or mass. It can only be transferred to other object via some interactions like collisions. So it has to be a. transferred to the box and that is the correct answer. </span>
Answer : The average speed of the sprinter is, 34.95 Km/hr
Solution :
Average velocity : It is defined as the distance traveled by the time taken.
Formula used for average velocity :

where,
= average velocity
d = distance traveled = 200 m
t = time taken = 20.6 s
Now put all the given values in the above formula, we get the average velocity of the sprinter.

conversion :
(1 Km = 1000m)
(1 hr = 3600 s)
Therefore, the average speed of the sprinter is, 34.95 Km/hr
Answer:
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
Explanation:
Total heat content of the fat = heat content of water +heat content of the lipids
Let it be Q
the Q= (mcΔT)_lipids + (mcΔT)_water
total mass of fat M= 0.63 Kg
Q= heat supplied = 100 W in 5 minutes
ΔT= 20°C
c_lipid= 1700J/(kgoC)
c_water= 4200J/(kgoC)
then,

solving the above equation we get
m= 0.46 kg
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
The refractive index for glycerine is

, while for air it is

.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:

where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case: