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2 years ago

11

Which wave has a longer period and how many times, the wave with a frequency of 7000Hz or the wave with a frequency of 21.000Hz?

1 answer:

Nuetrik [128]2 years ago

3 0

Answer:

I would say the answer is the wave of 21.000Hz

Explanation:

Because it has more frequency, and as more frequency you add, the time or longer period also increases.

You might be interested in

Where and when does the sun stay up for 24hours

DIA [1.3K]

Answer:

abisko, sweden

Explanation:

A bisco is the home to their eyes sky station an epic center for Aurora expanses and northern Sweden. During summer months, the Sun Bates to town and up to 24 hours of sunlight per day.

5 0

3 years ago

Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat

fomenos

The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².

<h3 /><h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

P₂=ρ₂gh₂

P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625 N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².

To learn more about the pressure refer to the link;

brainly.com/question/356585

#SPJ4

6 0

1 year ago

How many centimeters are in 3.50 feet?

Kazeer [188]

106.68 centimetres are in 3.50 feet

4 0

3 years ago

Which of the following lists the composition of the Earth’s outer core?

iris [78.8K]

Well,

The outer core of the Earth is mostly composed of iron and nickel.

The correct option is C.

5 0

3 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

2 years ago