Answer:
So we want to know what causes the electric charges to flow from one end of the battery to the other like the picture shows. In the battery we have a positive and a negative charge that are separated. They give an electric potential difference also known as voltage. The voltage causes the charge (electrons) to move from the negative pole to the positive pole. That is called the electron flow and that is what is what is actually happening in the electric circuit. The image shows that the current flows from the positive pole to the negative. That is called "conventional current" and it is a matter of convention. Both are caused by the difference in electric potential. So the correct answer is C.
We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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Answer:
v = 1.2 m/s
Explanation:
The wavelength of the waves is given as the horizontal distance between the crests:
λ = wavelength = 5.5 m
Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:
T = Time Period = 2(2.3 s) = 4.6 s
Now, the speed of the wave is given as:

where,
v= speed of wave = ?
f = frequency of wave = 
Therefore,

<u>v = 1.2 m/s</u>
Answer:
the average kinetic energy of water particles - C.
Answer:
Explanation:
General equation of the electromagnetic wave:
![E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%20E_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
where
Phase angle, 0
c = speed of the electromagnetic wave, 3 × 10⁸
wavelength of electromagnetic wave, 698 × 10⁻⁹m
E₀ = 3.5V/m
Electric field equation
![E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%203.5sin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)
Magnetic field Equation
![B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%20B_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
Where B₀= E₀/c

![B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)