What do these two changes have in common?

A student observes that it is hard to hear music underwater in a pool. They state that the sound is always muffled. They

s344n2d4d5 [400]

Answer:

FALSE

Explanation:

The answer is false.

The speed of the sound in water is faster when compared to the speed of sound in air. This is because, the particles in air is loosely packed and are far from each other as compared to water or liquid.

The water particles are close to each other than air particles, so water particles are able to transmit the vibrations of the sound faster than the air particles.

Therefore sound waves travels faster in water than in air.

5 0

3 years ago

Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.

Tcecarenko [31]

The energy of a photon is given by
$E=hf$
where
$h=6.6 \cdot 10^{-34} Js$ is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is
$f=5.49 \cdot 10^{14}Hz$
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
$E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J$

8 0

4 years ago

15 points and brainliest to whoever helps me!

Juliette [100K]

Answer:

The correct answer is false, a plant dying after being exposed to poison is not a physical change nor is it a physical property.

Explanation:

6 0

3 years ago

A rock is dropped from a height of 3.4 m. How much time does it take to hit

siniylev [52]

Answer: 33.32 s

Explanation: gravity =9.8m/s2 which means that 3.4mx9.8m/s2=33.32s

I hope this helped ! Sorry if it’s wrong :)

6 0

3 years ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration is given by;

α = (ωₓ - ωₐ) / T₁

α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

8 0

4 years ago