Answer:
Alternating
Explanation:
It is alternating because it is easy to distribute long distance.
Direct current is found in batteries and have large voltage drop over distance.
The Answer Is Lowering the amount of reactants
Answer: 4.7m/s²
Explanation:
According to newton's first law,
Force = mass × acceleration
Since we are given more the one force, we will take the resultant of the two vectors.
Mass = 2.0kg
F1+F2 = (3i-8j)+(5i+3j)
Adding component wise, we have;
F1+F2 = 3i+5i-8j+3j
F1+F2 = 8i-5j
Resultant of the sum of the forces will be;
R² = (8i)²+(-5j)²
Since i.i = j.j = 1
R² = 8²+5²
R² = 64+25
R² = 89
R = √89
R = 9.4N
Our resultant force = 9.4N
Substituting in the formula
F = ma
9.4 = 2a
a = 9.4/2
a = 4.7m/s²
Therefore, magnitude of the acceleration of the particle is 4.7m/s²
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.