A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.
<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>
Answer:
No, its not reasonable.
Explanation:
The substance that is to be dissolved is known as solute. The substance that is dissolving is known as solvent.
The amount of solvent in the mixture should be greater than that of solute.
Suppose we are taking a solvent in a beaker and we are continuously adding solute in it. Initially the solute dissolve quickly. At some point the solute stops dissolving in the solvent. This is known as saturation point of the solvent. After saturation point if solute is added further it does not dissolve in the solvent.
So, its not possible to dissolve 12.8 g of one substance in 11 g of another.
Answer: 1.65m
Explanation:
Refractive index in terms of the depth of liquid is the ratio of the real depth to the apparent depth of the liquid i.e Refractive index =Real depth/apparent depth
Refractive index of water given = 1.33
Real depth is the measure of how deep is the liquid while apparent depth is the depth at the surface of the liquid.
Real depth = 2.2m
Apparent depth =?
Applying the formula above
Apparent depth =Real depth/refractive index
= 2.2/1.33
= 1.65m
Therefore, the circle of light that exits the surface of the water when that light shines in the middle of the night is 1.65m wide
grow alot of plants that's help to make more h2o
Answer:
b. 20 sec
Explanation:
y = y₀ + v₀ t + ½ g t²
0 = 0 + (100) t + ½ (-10) t²
0 = 100t − 5t²
0 = t (100 − 5t)
t = 0, t = 20
The body lands after 20 seconds.
