Answer:
.c. −160°C
Explanation:
In the whole process one kg of water at 0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.
heat lost by 1 kg of water at 0°C
= mass x latent heat
= 1 x 80000 cals
= 80000 cals
heat gained by ice at −160°C to form ice at 0°C
= mass x specific heat of ice x rise in temperature
= 1 x .5 x 1000 x 160
= 80000 cals
so , heat lost = heat gained.
They would be likely to be underweight. This is because the role of villi is to increase absorption of soluble molecules, they do this by increasing surface area for absorption to occur across.
If the person has less villi than normal in their small intestine, then the surface area will not be as large meaning there is less area for absorption to occur across so less soluble molecules will be absorbed.
Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
(1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
(2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:

Next, you replace for all variables:

hence, the current in the wire is 4.75A
Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q = 
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC