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kirza4 [7]
3 years ago
10

How are mechanical waves and electromagnetic waves alike?

Physics
2 answers:
miss Akunina [59]3 years ago
6 0

I got this answer from the internet

Mila [183]3 years ago
5 0
Answer: Electromagnetic waves can travel through a vacuum; mechanical waves require a medium. Electromagnetic waves are visible; mechanical waves are invisible. Electromagnetic waves always reflect; mechanical waves always refract.




Explanation: There are two types of waves:
Mechanical waves: these waves consist of oscillations of the particles in a medium. Therefore, they can only propagates if there is a medium. Examples of mechanical waves are sound waves
Electromagnetic waves: they consist of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave. All electromagnetic waves travel in a vacuum always at the same speed, the speed of light: . They are the only types of waves that do not require a medium to propagate, so they can also travel through a vacuum.
We can now analyze the different statements:
Both types of waves require a medium. --> FALSE. Electromagnetic waves do not require a medium.
Both types of waves have a frequency. --> TRUE. All waves are characterized by their frequency, which is the number of complete oscillations per second.
Both types of waves transmit matter. --> FALSE. Waves transmit energy, but not matter.
Both types of waves have a pitch --> FALSE. Pitch tells how we perceive the frequency of a sound wave: but electromagnetic waves are not sound waves, so this statement is false.
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You walk exactly 250 steps North, turn around, and then walk exactly 400 steps South. How far are you from your starting
german

Answer:

150 steps south

Explanation:

250 north 250 back to start then continue south for remainder of 400 steps. 150 south

3 0
3 years ago
Each of the following would cause an increase in blood pressure except __________.
kicyunya [14]

Answer:

an inhibitor of angiotensin II

Explanation:

Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.

8 0
3 years ago
A light bulb that has more current running through it will...
RoseWind [281]
C
It does not affect the brightess of a lightbulb
4 0
3 years ago
A physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup. Show
motikmotik

Answer:

Decrease in level by thermal contraction  = 0.0076 mm

Drops in level = 3 mm

Therefore The decrease in level is not due to thermal contraction

Explanation:

From linear expansivity,

L₂ = L₁ +(1 + αΔT).............................. equation 1

Where L₂ = final length (m), L₁ = initial length, α = linear expansivity of liquid, ΔT = change in temperature.

Where L₁ = level of coffee in the glass cup = V/(πd²)/4..................... equation 2

     L₁ = 4V/πd², where V= Volume of coffee, d = diameter of the cup,

    L₁ = 350cm³, d = 7.00cm, and    π = 22/7 = 3.143

applying these values in equation 2 above,

     ∴  L ₁=( 4×350)/(3.143×7²)

       L₁ = 9.09

α = γ/3  Where γ = Volumetric expansion coefficient of liquid. = 0.00005/°C

α = 0.00005/3 = 0.0000167/°C, ΔT = L₂ - L₁ = 45 -95 = -50 °C.

Applying these values in equation 1 above,

L₂ = 9.09 [1+{0.0000167×(-50)}]

L₂ = 9.09 [1 - 0.000833]

L₂ = 9.09(0.999167)

L₂ = 9.082cm

∴ ΔL = Decrease in length = L₂ - L₁ = 9.082 - 9.09 = 0.3 cm

   ΔL = 0.0076

Decrease in level =0.076 mm

From the above, the decrease in level is not due to thermal contraction.

As Thermal contraction is less than Drops in level.

4 0
3 years ago
A vector has the components Ax=−31m and Ay=41m. What is the magnitude of this vector? What angle does this vector make with the
Fudgin [204]

Answer:

Magnitude of vector A is  51.4 m

The angle of vector A with positive x-axis is θ= 127.1^o}

Explanation:

Given is  

A_{x}=-31m

A_{y\\=41m

Let A_{x} and A_{y\\ are components of a vector A,

The magnitude of vector A will be:

A=\sqrt{A_{x}^2 +A_{y}^2}

   =  \sqrt{(-31)^{2} +41^2}

A  = 51.4 m

we know

 θ1= tan^{-1} \frac{A_{y}  }{A_{x} }

   = tan^{-1}(-\frac{41}{31}  )

   =-52.9^{o

Vector A is present in 2nd quadrant as x-component of vector A is negative and y-component is positive.

For finding the angle of vector with positive x-axis,

θ=180-52.9

θ= 127.1^o}

6 0
3 years ago
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