Answer:
<em>The force is now 9 times the original force</em>
Explanation:
<u>Coulomb's Law
</u>
The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's formula is:

Where:

q1, q2 = the particles' charge
d= The distance between the particles
Suppose the distance is reduced to d'=d/3, the new force F' is:




The force is now 9 times the original force
Answer:
The total amount of heat needed will be
.
Explanation:
We will divide the calculation in two: First, the heat needed to melt the ice, and then the heat needed to warm the resulting liquid from 0°C to 37°C.



<em>i) </em>The fusion heat will be:

<em>ii)</em> The heat needed to warm the water from
to
will be:

So, the total amount needed will be the sum of these two results:
.
Answer: 4.27 x 10^-10 N to the left
Explanation: I just took this quiz
Answer:
22145.27733 ft
124984.76055 ft
Explanation:
The equation of pressure is

where,
=Atmospheric pressure = 800 mbar
k = Constant
h = Altitude = 35000 ft


Now


The altitude will be 22145.27733 ft


The elevation is 124984.76055 ft