Household current in a circuit cannot generally exceed 15 for safety reasons. What is the maximum amount of charge taht could fl
1 answer:
The current I is defined as the amount of charge Q flowing through a certain point in a time t:
which can be rewritten as
so, by using this formula we can calculate the maximum amount of charge that can flow in the circuit. In fact, the maximum current is I=15 A, while the time is
so the maximum amount of charge is
which can be rewritten as
so, by using this formula we can calculate the maximum amount of charge that can flow in the circuit. In fact, the maximum current is I=15 A, while the time is
so the maximum amount of charge is
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Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume
Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol
As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)
When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V = (0.2029 m³)*(10³ L/m³) = 202.9 L
Answer: 203 liters
Answer:
<em>3.15 N towards the positive x-axis</em>
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Explanation:
first charge has charge q1 = 10 μC = 10 x 10^-6 C
second charge has charge q2 = 20 μC = 20 x 10^-6 C
third charge has charge q3 = -30 μC = -30 x 20^-6 C
According to coulomb's law, force between two charged particle is given as
F =
Where
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.
Q is the magnitude of one charge
q is the magnitude of the other charge
is the distance between these two charges
For the force on q2 due to q1,
distance r between them = 0 - (-1.0) = 1 m
F = = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)
For the force on q2 due to q3,
distance between them = 2.0 - 0 = 2 m
F = = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)
Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>
Answer:
The magnitude of acceleration is reduced.
Explanation:
Force is defined as push or pull
The force is said to be<em> balance force </em>if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.
The force are to be<em> unbalanced force </em>if the force cause change in motion. ie the object has force greater than zero and has acceleration.
According to <em>Newton second law of motion </em>, acceleration depends on force acting on the object and mass of object.
F=ma
a=
When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.