Voltmeter is used to find the potential difference between two points.
We always connect it in parallel to the points where we need the potential difference.
Here in order to make the reading accurate we can increase the resistance of voltmeter so that it can not withdraw any current from the circuit.
D. March because it is just below the 1 million marker on the graph and it is the only one that low.
Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
<span><em>78% = nitrogen</em>
<em>21% = oxygen</em>
</span>
<em>%1 = noble gases</em>
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
