So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6
It is respiration. Cell burns glucose with oxygen to release energy, along with producing water and carbon dioxide.
Answer:
Eo = 9.796 x 10^2 N/C
Bo = 3.266 x 10^-6 T
Explanation:
Given
Wavelength λ = 633 nm
Diameter of the beam D = 1.0 mm
Power P = 1.0 mW
Solution
Radius of the beam r = D/2 = 0.5 mm = 0.0005 m
Area of cross section
![A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E%7B2%7D%20%5C%5CA%20%3D%203.15%20%5Ctimes%200.0005%5E%7B2%7D%5C%5CA%20%3D%207.58%20%5Ctimes%2010%5E%7B-7%7D%20%20m%5E%7B2%7D%5C%5C)
Intensity
![I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BP%7D%7BA%7D%20%5C%5CI%20%3D%20%5Cfrac%7B0.001%7D%7B7.85%5Ctimes%2010%5E%7B-7%7D%7D%20%5C%5CI%20%3D%201273.885%20%7BW%7D%2F%7Bm%5E%7B2%7D%20%7D)
Amplitude of Electric Field
![E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C](https://tex.z-dn.net/?f=E_%7Bo%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2I%7D%7B%20%5Cepsilon_%7Bo%7Dc%20%7D%20%7D%20%5C%5CE_%7Bo%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%5Ctimes%201273.88%7D%7B%208.85%20%5Ctimes%2010%5E%7B-12%7D%20%5Ctimes%203%20%5Ctimes%2010%5E%7B8%7D%20%7D%20%7D%5C%5CE_%7Bo%7D%20%3D%209.796%20%5Ctimes%2010%5E%7B2%7DN%2FC)
Amplitude of Magnetic Field
![B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T](https://tex.z-dn.net/?f=B_%7Bo%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%5Cmu_%7Bo%7DI%7D%7Bc%20%7D%20%7D%20%5C%5CB_%7Bo%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%5Ctimes%204%20%5Ctimes%20%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctimes%201273.88%7D%7B%20%203%20%5Ctimes%2010%5E%7B8%7D%20%7D%20%7D%5C%5CB_%7Bo%7D%20%3D%203.266%20%5Ctimes%2010%5E%7B-6%7D%20T)
Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Answer:
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Explanation: