distance from the Sun of 2.77 astronomical units or about 414 million km 257 million miles and orbiting period of 4.62 years
Answer:
d. H + O2 ------ H2O
that's the formation of water
Answer:
option E
Explanation:
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time

N is the diameter of the aperture and t is the time of exposure
now,


inserting all the values

N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E
The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
Learn more about current:
brainly.com/question/25537936
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It's C, with the 2/7/4/6 in front of each reactant and product.