So we want to know what changes inside the multimeter when we change the voltage range from 200 V to 20 V, by what factor and does it increase or decrease. What we want when trying to measure the voltage with a multimeter is that a minimal current passes trough the mulitmeter so when we change the voltage range, we decrease the resistance by a factor of 10 because the voltage is decreased by a factor of 10.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
The first thing you should know in this case is the following definition:
PV = nRT
Then, as the temperature is constant, then:
PV = k
Then, we have two states:
P1V1 = k
P2V2 = k
We can then equalize both equations:
P1V1 = P2V2
Substituting the values:
(1.25) * (101) = (2.25) * (P2)
Clearing P2:
P2 = ((1.25) * (101)) /(2.25)=56.11Kpa
answer:
the new pressure inside the jar is 56.11Kpa
V = d ÷ t --> bc d=vt
V = (76+54)÷(2+5) = 130÷7 = 18.57km/hr
