The phenomenon of inducing voltage by changing the magnetic field around a conductor will be Electromagnetic Induction. Option B is correct.
<h3>What is the Faraday law of electromagnetic induction?</h3>
According to Faraday's law of electromagnetic induction, the rate of change of magnetic flux link with the coil is responsible for generating emf in the coil to result in the flow of amount of current .
So in order to increase the current, we need to increase the EMF;
so we can increase it by;
1) Increasing the number of turns
2) Increase the area of the loop
3) By moving the magnet faster
Hence, option B is correct.
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Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
Explanation:
Let yellow ball be m1 = 0.5kg with u1 = 8 m/s and blue ball be m2 = 0.25 kg with u2 = - 4 m /s respectively.
After collision, blue ball travels 12 m/s.
<u>Using conservation of Linear Momentum</u> :
m1u1 + m2u2 = m1v1 + m2v2
0.5* 8 + 0.25 * - 4 = 0.5 * v1 + 0.25 * 12
v1 = 0 m/sec i.e. <u>Yellow ball comes to rest</u>.
I think it is the red giant
Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
