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OleMash [197]
3 years ago
11

Why are readings in voltage different in a simple circuit

Physics
1 answer:
sveta [45]3 years ago
4 0

Answer:

The voltage is different across each component, because in a series circuit, the current is constant as there is only one path for current to flow.

Explanation:

Hope it helps! :3

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What is the formula for Charles Law?
Vlada [557]

Answer:

Explanation:

V_{1} = first volume

V_{2} = second volume

T_{1} = first temperature

T_{2} = second temperature

4 0
3 years ago
S27253129 ,,, message me please, I can't ask you my homework question in the comments :c
Serjik [45]
What’s the problem what do u need help on
7 0
3 years ago
Find the ratio of effusion rates of hydrogen gas and krypton gas.
algol13

Answer:

RE of Hydrogen = 6.47 x RE of Krypton

Explanation:

Actually the correct formula for comparing rate of effusion (RE) of two gases is:

RE of Gas A

------------------- = √ ( Molar mass of B / Molar mass of A)

RE of Gas B

You can designate which of the two gases you have (hydrogen and krypton) will be your gas A and gas B. So for this particular problem, let us make hydrogen as gas A and Krypton as gas B. So the equation becomes:

RE of Hydrogen

------------------------- = √ (Molar mass of Krypton / Molar mass of Hydrogen)

RE of Krypton

Get the molar masses of Hydrogen and Krypton in the periodi table:

RE of Hydrogen

------------------------- = √ (83.798 g/mol / 2 g/mol)

RE of Krypton

RE of Hydrogen

------------------------- = 6.47  ====> this can also be written as:

RE of Krypton

RE of Hydrogen = 6.47 x RE of Krypton

It means that the rate of effusion of Hydrogen gas will be 6.47 faster than the rate of effusion of Krypton gas. With the type of question you have, it doesn't matter which gases goes on your numerator and denominator. What's important is that you show the rate of effusion of a gas with respect to the other. But if that's concerns you the most, then take the gas which was stated first as your gas A and the latter as your gas B unless the problem tells you which one will be on top and which is in the bottom.

3 0
3 years ago
Find the magnitude of the force needed to accelerate a 300 g mass with a⃗ = -0.205 m/s2 i^+0.700 m/s2 j
never [62]
F=ma
m=300g= 0.3kg
a= -0.205i +0.700j m/s2 = <-0205, 0.7>m/s2

Thus, F = 0.3<-0.205, 0.7> N
F = <-0.0615, 0.21> N
= -0.0615i N +0.2100j N
7 0
3 years ago
What is the magnitude of a point charge that produces a potential of -200V at a distance of 1.00 mm?
schepotkina [342]

Answer:

q=-2.22*10^{-11}C

Explanation:

The potential produces by a point charge is given by:

V=\frac{kq}{r}

Here, k is the Coulomb constant, q is the signed magnitude of the point charge and r is the distance between the charge and the point at which the electric potential is measured. Solving for q:

q=\frac{rV}{k}\\q=\frac{1*10^{-3}m(-200V)}{8.99*10^9\frac{V\cdot m}{C}}\\q=-2.22*10^{-11}C

5 0
3 years ago
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