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
1) 29.8 C
At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).
2) 6.2 C
The temperature change of the water is given by the difference between its final temperature and its initial temperature:
where
Substituting into the formula,
And the positive sign means that the temperature of the water has increased.
3) -40.6 C
The temperature change of the metal is given by the difference between its final temperature and its initial temperature:
where
Substituting into the formula,
And the negative sign means the temperature of the metal has decreased.
Density is given as
now we have to convert this density into 
now we have
Answer:
Explanation:
Mass of ball Is m=96.1g=0.0961kg
Height above spring is 59.1cm
L=0.591m
Extension of the spring is 4.75403cm
e=0.0475403m
Then the distance the ball traveled is H=L+e
H=0.591+0.0475403
H=0.6385403m
Then, the potential energy of the ball is given as
P.E=mgh
P.E=0.0961×9.81×0.6385403
P.E=0.602J
From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another
Then, the P.E is transferred to the work done by the spring
Then, Work done by spring is given as
W=½ke²
W=P.E=½×k×0.0475403²
0.602=½×k×0.0475403²
k=0.602×2/0.0475403²
k=532.72N/m
The spring constant is 532.72 N/m
Answer:
The distance from the charge is 3.35 m.
Explanation:
Given that,
Electric potential, V = 635 V
Magnitude of electric field, E = 189 N/C
We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :
d is the distance from charge
So, the distance from the charge is 3.35 m. Hence, this is the required solution.
