A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time
1 answer:
Answer:
B. The current increases.
Explanation:
As we know that rate of flow of charge through the conductor is known as electric current
So we have
here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased
so we can say that the ratio of charge and time will increase
so here we have
So correct answer will be
B. The current increases.
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Refer to the diagram shown below.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.