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3 years ago

8

If the force on the tympanic membrane (eardrum) increases by about 1.4 N above the force from the atmospheric pressure, the memb

rane can be damaged. When you go scuba diving in the ocean, below what depth (in feet) could damage to your eardrum start to occur? Considering that your eardrum is 7.4 mm in diameter. Consider the density of sea water to be 1.03*10^3 kg/m^3.

1 answer:

Sveta_85 [38]3 years ago

6 0

Answer:

10.58 ft

Explanation:

Force, F = 1.4 N

Diameter of membrane = 7.4 mm

radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m

Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2

Density, d = 1.03 x 10^3 kg/m^3

Pressure at depth, P = h x d x g

Let h be the depth.

Pressure = force / Area

h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)

h = 3.225 m = 10.58 ft

Thus, the depth of water is 10.58 ft.

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Answer:

Both

Explanation:

The S.I. unit of pressure is newton/meter square or pascal as both represent the same dimensional value.

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2 years ago

When an excited electron falls from n = 3 to n = 2, a red radiation in the Balmer series emitted. What is the frequency of this

vichka [17]

Answer:

The frequency of this photon is $4.57\times10^{14}\ Hz$

(D) is correct option.

Explanation:

Given that,

Excited states,

$n=3$

$n=2$

We need to calculate the wavelength

Using formula for energy

$E=13.6(\dfrac{1}{n^2}-\dfrac{1}{m^2})$

$E=13.6(\dfrac{1}{4}-\dfrac{1}{9})$

$E=1.888\ eV$

$E=1.888\times1.6\times10^{-19}\ J$

$E=3.0208\times10^{-19}\ J$

We need to calculate the frequency

Using formula of frequency

$E=hf$

$f=\dfrac{E}{h}$

Where, E =energy

$f=\dfrac{3.0208\times10^{-19}}{6.63\times10^{-34}}$

$f=4.57\times10^{14}\ Hz$

Hence, The frequency of this photon is $4.57\times10^{14}\ Hz$

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2 years ago

Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\

andreev551 [17]

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

θ₂ = 68.0-22.2 = 45.8º

I = I₂ cos² θ₂

I₂ = I / cos₂ θ₂

I₂ = 75.5 / cos² 45.8

I₂ = 155.3 W

We repeat for First and second polarizer

I₂ = I₁ cos² θ₁

I₁ = I₂ / cos² θ₁

I₁ = 155.3 / cos² 22.2

I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

I₁ = I₀ / 2

I₀ = 2 I₁

I₀ = 2 181.2

I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

I₁ = 181.2 W

The intensity at the exit is

I₂ = I₁ cos² θ₂

I₂ = 181.2 cos² 68.0

I₂ = 25.4 W

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3 years ago

Tripling the displacement from equilibrium of an object in simple harmonic motion will bring about a change in the magnitude of

Anni [7]

Answer:

acceleration will be tripled.

Explanation:

We know, when an object is performing Simple harmonic motion, the force

experience by it is directly proportional to its displacement from its mean position.

Also, F = ma , therefore, acceleration is also proportional to its displacement .

Now, F = kx

Therefore, $a=\dfrac{k\ x}{m}$

If we triple the displacement i.e, 3x.

Acceleration $a'=\dfrac{k(3x)}{m}=3a.$

Therefore, acceleration is also tripled.

Hence, this is the required solution.

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3 years ago

How much heat is released as a 5.89 kg block of aluminum cools from 462 °C to 315 °C. The specific heat capacity of aluminum is

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779,247 J is the amount of heat released.

<u>Explanation:</u>

The equation that gives the amount of heat supplied is

$E \text { or } Q=m \times c_{A l} \times \Delta T$

Where,

E or Q – amount of heat supplied

m- mass

$c_{A l}$ - specific heat capacity of Aluminium

$\Delta T$ – temperature variation caused by heat change

The amount of heat energies that causes the temperature to vary $1^{\circ} \mathrm{C}$ or 1 K per kg of the material is known as specific heat capacity (c)

Here, Given data:

$c_{A l}$ – 0.900 J/g.K

To convert gram into kilogram, multiply and divide by $10^{3}$ , we get

Specific heat capacity of aluminum = 900 J/kg. K

m – 5.89 kg

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By substituting the known values in the above equation, we get

$E \text { or } Q=5.89 \times 900 \times 147=779,247 \mathrm{J}$

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3 years ago

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