150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit
1 answer:
The current in the circuit is 5 A
Explanation:
The intensity of current is given by the equation:
where
I is the current
q is the amount of charge passing through a given point of the circuit in a time interval of t
For the cell in this problem, we have
q = 150 C is the charge
t = 30 s is the time interval
Substituting into the equation, we f ind
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Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
Answer:
(a) 1500 m
(b) 2827.43m
Explanation:
Given that the time for one cycle of the swing is 1 s
The radius of the swing, R= 3.0m
The angle covered, \theta, by each swing is a quarter of the circle. i.e.
Speed of running of women =5 m/s.
Time of running = 5 minutes= 5 x 60 secondes= 300 s.
(a) As distance= (speed) x (time)
So, the required distance= 5 x 300 m= 1500 m.
(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.
As arc length, , where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.
So, the distance covered by the child in 1 swing .
In 1 cycle, there are 2 swings, so distance covered in 1 cycle .
Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.
So, the total distance covered by the child
.
