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:
Temperature at the exit = 
Explanation:
For the steady energy flow through a control volume, the power output is given as

Inlet area of the turbine = 
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume 
as


for Ideal gasses, the enthalpy change can be calculated using the formula

hence we have


<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have 
Hence, the temperature at the exit = 
Answer:
The current is
Explanation:
From the question we are told that
The radius of the loop is 
The earth's magnetic field is 
The number of turns is 
Generally the magnetic field generated by the current in the loop is mathematically represented as

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

=> 
Where
is the permeability of free space with value 

=> 

Answer:B.As the north pole of the electromagnet nears the south pole of the permanent magnet, the current reverses and the poles of the magnets then repel.
Explanation: