Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer: a network of several radio telescopes wired together
Explanation:
A radio interferometer combines signals of several radio telescopes which are used in astronomical observations simultaneously to simulate a discretely-sampled single telescope of very large aperture
Interferometer, an instrument that uses the interference patterns formed by waves to measure certain characteristics of the waves themselves or of materials that reflect, refract, or transmit the waves. Interferometers can also be used to make precise measurements of distance.
<h2>
Answer:</h2>
In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:
So the average power is:
But:
So:
![P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%28%5Cfrac%7B1%2Bcos%282%5Comega%20t%29%7D%7B2%7D%20%29dt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5Cintop_%7B0%7D%5E%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%2B%5Cfrac%7Bcos%282%5Comega%20t%29%7D%7B2%7D%5Ddt%20%5C%5C%5C%5CP%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7BT%7D%5B%5Cfrac%7B1%7D%7B2%7D%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B4%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2T%7D%5B%28t%29%5Cright%7C_0%5ET%20%2B%5Cfrac%7Bsin%282%5Comega%20t%29%7D%7B2%5Comega%7D%20%5Cright%7C_0%5ET%5D%20%5C%5C%20%5C%5C%20P%3D%5Cfrac%7Bv_%7Bm%7Di_%7Bm%7D%7D%7B2%7D)
In terms of RMS values:
<span>Mendeleev observed that tellurium has chemical properties like other elements in its group, and he did not know that neutrons cause the greater atomic mass is the most likely reason.</span>
