Answer:
The voltage is
Explanation:
From the question we are told that
The time that has passed is 
Here
is know as the time constant
The voltage of the power source is 
Generally the voltage equation for charging a capacitor is mathematically represented as
![V = V_b [1 - e^{- \frac{t}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Cfrac%7B%5Ctau%7D%7B2%7D%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\tau}{2\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Ctau%7D%7B2%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{1}{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B1%7D%7B2%7D%20%7D%5D)
=>
The spectrum of light from the moon should very strongly resemble the spectrum of sunlight. The reason is that any light from the moon started out from the sun. Any difference in their spectra is only due to the moon absorbing more of some wavelengths and less of others. But since the moon appears colorless gray, we don't expect any particular colors to be strongly absorbed, otherwise the moon would look to be the colors of the light that's left.
Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:

b) The escalar product a·b is:

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
![(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5Ccdot%20b%20%3D%20%5B%288.6%20%2B%209.3%29i%20%2B%20%285.1%20%2B%209.5%29j%5D%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20%2817.9i%20%2B%2014.6j%29%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20305.17)
d) The component of a along the direction of b is:

I hope it helps you!
In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then
1/Rt = 1/7 + 1/49
1/Rt = 1/6.125 = 1/ 49/8
Rt = 49/8 <span>Ω
The total resistance hence is </span>49/8 Ω