Answer: q2 = -0.05286
Explanation:
Given that
Charge q1 = - 0.00325C
Electric force F = 48900N
The electric field strength experienced by the charge will be force per unit charge. That is
E = F/q
Substitute F and q into the formula
E = 48900/0.00325
E = 15046153.85 N/C
The value of the repelled second charge will be achieved by using the formula
E = kq/d^2
Where the value of constant
k = 8.99×10^9Nm^2/C^2
d = 5.62m
Substitutes E, d and k into the formula
15046153.85 = 8.99×10^9q/5.62^2
15046153.85 = 284634186.5q
Make q the subject of formula
q2 = 15046153.85/ 28463416.5
q2 = 0.05286
Since they repelled each other, q2 will be negative. Therefore,
q2 = -0.05286
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Answer:
Explanation:
In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:
In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:
The initial potential energy of the spring is given by the equation:
the Kinetic energy of the block is then given by the equation:
so we can now set them both equal to each other, so we get:
This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:
so now we can solve this for the final velocity, so we get:
Answer:
GFCI outlets are found in wet areas.
GFCI outlets prevent electrocution if you are touching a wet appliance.
Answer:
Explanation:
- For diagram refer the attachment.
It is given that five cells of 2V are connected in series, so total voltage of the battery:
Three resistor of 5, 10, 15 are connected in Series, so the net resistance:
According to ohm's law:
On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 we get:
The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>