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2 years ago

We can block light by placing obstacles in its path, but it’s much more difficult to block sound. Why?

Physics

2 answers:

Flauer [41]2 years ago

8 0

<h3>Answer;</h3>

B. Sound waves can move through various mediums.

<h3>Explanation;</h3>

A wave is a transmission of a disturbance from one point to another. It involves transmission of energy from the source to another point.
A wave may or may not require a medium for transmission. Waves that require medium for transmission are called mechanical waves and those that do not require medium for transmission are known as electromagnetic waves.
Light is an example of electromagnetic wave therefore it can be blocked when obstacles are placed on its path. However, for the case of sound waves, they can not be blocked by an obstacle as sound would travel through the medium used to block.

mamaluj [8]2 years ago

6 0

Answer:

B. Sound waves can move through various mediums.

Explanation:

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gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an

Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

Explanation:

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0

2 years ago

Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0

2 years ago

A cup of hot coffee initially at 95 degrees C cools to 80 degrees C in 5 min while sitting in a room of temperature 21 degrees C

Oksana_A [137]

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

Given;

The initial temperature of the coffee T₀ = 95 °C

The temperature of the room = 21°C

Let T be the temperature at time of cooling t in mins

According to Newton's law of cooling;

$\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}} = \int\limits kdt\\\\Log(T-21) =kt + Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453$

When the temperature is 50 °C, the time t in min is calculated as;

$T -21 = 74e^{-0.0453t}\\\\50 -21 = 74e^{-0.0453t}\\\\29 = 74e^{-0.0453t}\\\\\frac{29}{74} = e^{-0.0453t}\\\\0.39189 = e^{-0.0453t}\\\\ln(0.39189 ) = {-0.0453t}\\\\-0.93677 = {-0.0453t}\\\\t = \frac{-0.93677}{-0.0453}\\\\ t = 20.68 \ mins$