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dem82 [27]
3 years ago
11

What is the speed of a wave with a period of 25 sec/cycle and a wavelength of 15 m/cycle?please help me !

Physics
1 answer:
Nuetrik [128]3 years ago
5 0
The answer of this question is 0.6m/s
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Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
2 years ago
How many neutrons does element X have if it’s atomic number is 28 and it’s mass number is 81?
labwork [276]

Well let’s put it this way. To find the neutrons you subtract the atomic atomic Nuremberg from the atomic mass. So

Mass=81-Number=28

81-28=53

Final answer is 53.

5 0
3 years ago
Which vector correctly indicates the direction of centripetal acceleration experienced by the car?
gogolik [260]

The answer is C.

I hope this helped!

5 0
3 years ago
Read 2 more answers
How is a revolution related to a year
nikdorinn [45]
Revolution means orbiting around another body. 
<span>A year is the time for a planet to complete one orbit around the Sun. 100% sure

</span>
4 0
3 years ago
Read 2 more answers
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
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