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3 years ago

Which statement correctly describes both sound and light waves?

Physics

1 answer:

olga55 [171]3 years ago

6 0

Answer:

Option C. Light waves can travel through space, sound waves cannot.

Explanation:

Light is an electromagnetic wave which requires no Meduim for propagation. However, sound wave requires a medium for propagation. Since space is empty, sound can not travel through it.

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Question 12 (1 point) Question 12 Unsaved

Bezzdna [24]

Slope is your answer

7 0

3 years ago

Which one of the following is not equivalent to 2.50 miles?

lisabon 2012 [21]

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)$

$x = 13200\,ft$

$x = 1.320\times 10^{4}\,ft$ (Choice A)

mi - in

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)$

$x = 158400\,in$

$x = 1.584\times 10^{5}$ (Choice B)

mi - km

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)$

$x = 4.025\,km$ (Different from Choice C)

mi - yd

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft} \right)$

$x = 4400\,yd$

$x = 4.40\times 10^{3}\,yd$ (Choice D)

mi - cm

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)$

$x = 402500\,cm$

$x = 4.025\times 10^{5}\,cm$ (Choice E)

Choice C is not equivalent to 2.50 miles.

6 0

3 years ago

PlzHELP

Naddika [18.5K]

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.

For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me.

https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php

Hope I helped.

8 0

3 years ago

The average Intensity of the electromagnetic field due to sunlight at the surface of the earth is 1400 W/m^2. Calculate the maxi

Ann [662]

Answer:

1027 N/C

3.42 x 10⁻⁶ T

Explanation:

I = Intensity of electromagnetic field = 1400 W/m²

E₀ = Maximum value of electric field

Intensity of electromagnetic field is given as

I = (0.5) ε₀ E₀² c

1400 = (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²

E₀ = 1027 N/C

B₀ = maximum value of magnetic field

using the equation

E₀ = B₀ c

1027 = B₀ (3 x 10⁸)

B₀ = 3.42 x 10⁻⁶ T

4 0

3 years ago

Which of the following properties must be known in order to calculate the amount of heat needed to melt 1.0 kg of ice at 0ºC?

meriva

If you're only melting ice ... turning solid ice at 0°C into liquid water at 0°C ...
then you only need to know the latent heat of fusion for water.

That's exactly what it means ... the amount of energy involved when a gram
of H₂O changes in either direction between liquid and solid, with no change
in temperature.

7 0

3 years ago