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NeTakaya
3 years ago
13

How do you know that waves sent from the sun to earth are not mechanical waves? Explain

Physics
1 answer:
Anna71 [15]3 years ago
3 0
Mechanical waves require matter to travel medium. If there is no substance or matter to travel through, mechanical waves cannot propagate.

Since space is mostly vacuum, there is no medium for waves to travel through: waves sent by the sun (electromagnetic waves) are not mechanical waves
You might be interested in
Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle 0 between th
erik [133]

Based on the mass of the bob and the angle between the cord and the vertical, the pendulum length is 0.50m.

The maximum kinetic energy can be found to be 9.42 x 10⁻⁴J.

<h3>What is the pendulum length?</h3>

This can be found as:

= g-force / w²

Solving gives:

= 9.8 / 4.43²

= 0.4998 m

= 0.50 m

<h3>What is the maximum kinetic energy?</h3>

This can be found as:

= 0.5 × m × w² × A²

Maximum kinetic energy is:

= 0.5 × 60 × 10⁻³ × (4.43 × 0.4998 x 0.08 rad)²

= 9.42 x 10⁻⁴J

Find out more on maximum kinetic energy at brainly.com/question/24690095.

5 0
2 years ago
What is the inverse square law and how does it relate to gravity?
Nesterboy [21]

Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

7 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
The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s
alex41 [277]
Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
8 0
3 years ago
Riding in a car, you suddenly put on the brakes. As you experience it inside the car, do Newton's law apply? Do they apply as se
alisha [4.7K]

Answer with Explanation:

Newton's laws are applicable for inertial frames of reference which is a frame which is not accelerating when seen from the observer standing on earth.

For the person as he presses the brakes his frame is a decelerating frame of reference hence he cannot apply the newtons laws of motion as they are in their original form but if he analyses the motion he has to apply a correction known as  pseudo-force on the object he is analyzing. Pseudo Force has no basis in newton's laws but are a correction that needs to be applied if he wishes to analyse the motion from non inertial frame of reference

While as a person standing on earth outside the car since his frame is an inertial frame of reference he can apply newton's laws of motion without any correction.  

3 0
3 years ago
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