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2 years ago

Light from the sun travels in electromagnetic waves. Which of these statements describe electromagnetic waves?

Physics

1 answer:

kupik [55]2 years ago

7 0

Complete Question:

Light from the sun travels in electromagnetic waves. Which of these statements describe electromagnetic waves? Choose the 3 that apply.

Group of answer choices.

a) consist of transverse wave patterns.

b) can move through a vacuum.

c) caused by a vibration in a medium.

d) consist of longitudinal wave patterns.

e) requires a medium, such as air, to travel.

f) made of vibrating electric and magnetic fields.

Answer:

a) consist of transverse wave patterns

b) can move through a vacuum

f) made of vibrating electric and magnetic fields

Explanation:

Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.

Generally, the most commonly used electromagnetic wave technology in telecommunications is light and radio waves. Light from the sun travels in electromagnetic waves.

The statements which describe electromagnetic waves are;

I. It consist of transverse wave patterns.

II. It can move through a vacuum.

III. It is made of vibrating electric and magnetic fields.

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You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You disc

FrozenT [24]

Answer:

The balloon hit the ground with velocity -15.34 m/s

Explanation:

Lets explain how to solve the problem

You found that the best height to pitch a water balloon in order for it to

burst when it hits the ground is 12 meters.

We consider that the 12 meters is the maximum height, so the velocity

at this height is zero.

To find the velocity when the balloon hits the ground lets use the rule

v² = u² + 2gh, where v is the final velocity, u is the initial velocity, g is

the acceleration of gravity and h is the height.

u = 0 , h = 12 m , g = 9.8 m/s²

Substitute these values in the equation above

v² = 0 + 2(9.8)(12)

v² = 235.2

Take square root for both sides

v = ± $\sqrt{235.2}$

The velocity is downward, then it's a negative value

Then v = -15.34 m/s

The balloon hit the ground with velocity -15.34 m/s

6 0

2 years ago

A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature

Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

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3 years ago

Nick is so excited to Trick-or-Treat, he RUNS down the street to the house that has the BIG candy bars! He ran 203 meters. It to

aleksandr82 [10.1K]

對不起，我沒有看到照片，我看到的都是黑色的東西。對不起。

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2 years ago

A 10.0 Ω lightbulb is connected to a 12.0 V battery. (a) What current flows through the bulb? (b) What is the power of the bulb?

andreev551 [17]

The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

<h3>What current flows through the bulb as well as the power of the bulb?</h3>

From ohm's law; V = I × R

Where V is the voltage, I is the current and R is the resistance.

Also, Power is expressed as; P = V × I

Where V is voltage and I is current.

Given that;

Resistance R = 10.0 ohms
Voltage V = 12.0V
Current I = ?
Power P = ?

First, we determine the current flow through the bulb.

V = I × R

12.0V = I × 10.0 ohms

I = 12.0 ÷ 10.0

I = 1.2A

Next, we determine the power of the bulb.

P = V × I

P = 12.0V × 1.2A

P = 14.4 Watts

Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

Learn more about Ohm's law here: brainly.com/question/12948166

#SPJ1

6 0

1 year ago

Newton’s law of gravitation says that gravity is a mutually attractive force. Explain the following observation: A small object

Airida [17]

Answer: See below

Explanation:

The Earth attracts the falling object with the same intensity of gravity as the object attracts the Earth, according to Newton's law of gravitation. The displacement of the two bodies, however, is inversely proportional to their respective masses.

Example: The Earth attracts a ball that falls 3 metres from the ground, even though the ball's mass is insignificant in comparison to the Earth's. Similarly, the ball draws the Earth with the same power, but the Earth's mass is enormously more than the ball's. As a result, the Earth collides with a billionth of a millimetre ball (or even less). Restart the Earth's descent on the ball you'll never see again.

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4 0

2 years ago

Read 2 more answers