<h2>Answer:</h2>
A series circuit occurs when the elements are connected along a simple path so the same current flows through all the elements. On the other hand, a parallel circuit occurs when there are two or more paths for the electricity to flow. The diagram are shown in the Figure below. We have chosen a source and resistors to illustrate this problem.
From the information given and if the question is complete then;
Absolute temperature is the temperature in Kelvin
To convert degree Celsius to kelvin we normally add 273
that is Kelvin = deg Celsius + 273
Thus since we have been given that the air was at -70 degrees celcius;
then; - 70° C + 273 = 203 K
Therefore; the absolute temperature is 203 K
E - impossible to determine based on the given information
Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
---- 
--- 
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
--- Coulomb's law
Substitute 2 for r1


The new force (F2) is

Substitute 4 for r2



Substitute 


The new force is 1/4 of the previous force.
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5