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Neporo4naja [7]
3 years ago
7

Calculate the period of wave whose frequency is 50Hz​

Physics
2 answers:
harkovskaia [24]3 years ago
8 0

Answer:

0.02 sec

here.

frequency=50 Hz

we know that time period = 1/f

=1/50

=0.02sec

miss Akunina [59]3 years ago
5 0

Explanation:

time period= 1/frequency.

= 1/50 = 0.02 second.

hope this helps you.

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A stone is thrown from a cliff and lands in the sea. Air resistance is negligible. Which statement is correct whilst the stone i
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Since the stone is travelling, the displacement cannot be constant. This leaves the last 2 choices as possibilities. Now consider the acceleration that can act on the stone. Acceleration due to gravity is the only one. That means the horizontal component of the stone has to be constant since there is no acceleration in the horizontal direction.
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3 years ago
What are common features of both the MMPI-2 and CPI? Select all that apply.
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Answer:     D

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MMPI - 2

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8 0
3 years ago
A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
3 years ago
Which of the following is a chemical equation that accurately represents what happens when a sulfur and oxygen are produced from
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It's c................
5 0
3 years ago
In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo
Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find  the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Where, s = distance

t = time

a = acceleration

Put the value into the formula

60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2

a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}

a=0.0244\ m/s^2

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0
3 years ago
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