Answer:
the reading on the scale is 158.5 mm.
Explanation:
Given;
upper fixed point of the temperature scale, x₁ = 260 mm
lower fixed point of the temperature scale, x₂ = 50 mm
upper temperature scale, T₁ = 212 °F
lower temperature scale, T₂ = 32 °F
thermometer reading, t = 125 °F, let the reading on the scale = x
Interpolate as follows to determine the value of "x"
Therefore, the reading on the scale is 158.5 mm.
The best object to show the Earth's true appearance is something round and smooth. Since the earth is only slightly oblate, it doesn't matter based on its size, and you would use a pool ball as the perfect scale model, because it is very round and smooth.
It’s Li. Group one elements are more reactive. Reactivity decreases down the group and also decreases across the periods
Answer:
a) Q1= Q2= 11.75×10^-6Coulombs
b) Q1 =15×10^-6coulombs
Q2 = 38.75×10^-6coulombs
Explanation:
a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as
1/Ct = 1/C1 + 1/C2
Given C1 = 3.00 μF C2 = 7.75μF
1/Ct = 1/3+1/7.73
1/Ct = 0.333+ 0.129
1/Ct = 0.462
Ct = 1/0.462
Ct = 2.35μF
V = 5.00Volts
To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance
Q = 2.35×10^-6× 5
Q = 11.75×10^-6Coulombs
Since same charge flows through a series connected capacitors, therefore Q1= Q2=
11.75×10^-6Coulombs
b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2
C = 3.00 μF + 7.75 μF
C = 10.75 μF
For 3.00 μF capacitance, the charge on it will be Q1 = C1V
Q1 = 3×10^-6 × 5
Q1 =15×10^-6coulombs
For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5
Q2 = 38.75×10^-6coulombs
Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.