Can someone please help, ty!!<br> (Will mark brainliest)

Radio waves of frequency 1.667 GHzGHz arrive at two telescopes that are connected by a computer to perform interferometry. One p

SVETLANKA909090 [29]

Explanation:

The frequency of radio waves is 1.667 GHz

One portion of the same wave front travels 1.260 mm farther than the other before the two signals are combined.

There are two conditions for interference either constructive or destructive.

For constructive interference , the path difference is n times of wavelength and for destructive interference, the path difference is (n+1/2) times of wavelength

We can find wavelength in this case as follows :

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1.667\times 10^9}\\\\\lambda=0.1799\ m$

If we divide path difference by wavelength,

$\dfrac{\delta}{\lambda} =\dfrac{1.26}{0.1799}\\\\\dfrac{\delta}{\lambda} =7\\\\\delta =7\lambda$

It means that the path difference is 7 times of the wavelength. it means the two waves combine constructively and the value of m for the path difference between the two signals is 7.

5 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago

Which atomic number is the threshold value below which fusion may occur?

Luda [366]

Answer:

The atomic number 26(iron) is the threshold value below which the fusion might occur.

Explanation:

Nuclear fusion is a reaction in which two or more nuclei are combined to form one or more different atomic nuclei and subatomic particles.

Energy released in a fusion reaction is because of a key feature of nuclear matter called the binding energy which is a measure of the efficiency with which its constituent nucleons are bound together.

As we go up in atomic number, the energy released per nuclei goes down until it hits a minimum which is for atomic number 26 (iron) and fusion is not possible.

7 0

3 years ago

Read 2 more answers

How many watt-hours does a 80 watt light bulb use if it runs for 3.5 hours?

coldgirl [10]

Answer:

5 I Think

Explanation:

3 0

3 years ago

Numbered 8 nowwwwwww

amm1812

Torque = r x F

|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)

distance of fulcrum = torque / Force = 90/600 m = .15 m.

7 0

3 years ago

Can someone please help, ty!! (Will mark brainliest)