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3 years ago

What is the function of hammer and anvil in human ear

Physics

2 answers:

polet [3.4K]3 years ago

6 0

Answer:

The Hammer, Anvil, and Stirrup magnify the oscillations and relay this information to the chamber of the inner ear.

MArishka [77]3 years ago

4 0

Answer:

Vibrations of the eardrum are in turn translated into oscillations of tiny bones (ossicles) found within the middle ear. The Hammer, Anvil, and Stirrup magnify the oscillations and relay this information to the chamber of the inner ear.

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Professor Frank Nabarro insists that all senior physics majors take his notorious physics aptitude test. The test is so tough th

yulyashka [42]

Answer:

0.5504

Explanation:

We can use Bay's theorem to solve this question

P( No physics career/fail) = P(No Physics career and fail)/P(Fail)

= P( Fail/ No physics career)×P( no physics career)/P( fail)

= $\frac{0.3\times1}{(0.7\times0.35)+(0.3\times1)}$

=0.5504

8 0

3 years ago

How much energy can be stored in a spring with k = 470 n/m if the maximum possible stretch is 17.0 cm ?

MrRissso [65]

The strain energy stored in a linear spring is
SE = (1/2)*k*x²
where
k = the spring constant
x = the extension (or compression) of the spring

Given:
k = 470 N/m
x = 17.0 cm = 0.17 m

Therefore
SE = 0.5*(470 N/m)*(0.17 m)² = 6.7915 J

Answer: 6.8 J (nearest tenth)

6 0

4 years ago

i need help please. this is for physics but everything i search for related to this comes up as chemistry

Annette [7]

The car tyre contains air initially at a pressure of 195 kPa after travelling several km the temperature of the air inside a car tyre rises from 30 to 70°C if the tyre is rigid and does not expand then the new pressure inside the tyre would be 220.74 kPa.

<h3>What is pressure?</h3>

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

By using the Charles law for gases which states that the volume of the gas remains constant then the pressure of the gas is directly proportional to the temperature.

As given in the problem the tyre is rigid and does not expand this means the volume of the tyre remains constant.

The mathematical expression for Charles's law is as follows

P₁/P₂ = T₁/T₂

First, we have to change the temperature from degree Celcius to the kelvin temperature scale

K = 273 + C

where k is the temperature in kelvin and the C is degrees of Celcius

Initially, the temperature was 30° C

T₁ = 273 + 30

T₁ = 303 K

Then after travelling the temperature of the air inside a car tyre rises from 30 to 70°C

T₂= 273+ 70

T₂ =343 K

The car tyre contains air initially at a pressure of 195 KPa

P₁ = 195 kPa

Lets us take the final pressure of the air would be P₂

By substituting the values in the formula

P₁/P₂ = T₁/T₂

195/P₂ = 303/343

P₂ = 220.74 kPa

Thus, the new pressure inside the tyre would be 220.74 kPa.

Learn more about pressure learn more

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#SPJ1

7 0

2 years ago

In a pulley system, two blocks are connected by a rope as shown below. The coefficient of static friction between block A and th

umka21 [38]

Hi there!

We can begin by doing a summation of forces for each block.

Block A:

This block has the force of tension (in direction of acceleration +) and static friction (opposite direction -) acting on it. Thus:

$\Sigma F_A = T - F_s\\\\m_Aa = T - \mu m_Ag$

Block B:

This block has the force of tension (opposite of acc. -) and gravity (in direction of acc +), working on it.

$\Sigma F_B = m_Bg - T\\\\ m_Ba = m_Bg - T$

Add both of the expressions and solve for the maximum mass of Block B.

$\Sigma F = m_Bg - T + T - \mu m_Ag\\\\a(m_A + m_B) = m_Bg - \mu m_Ag$

To find the minimum value, we can set a = 0, so:

$0 = m_Bg - \mu m_Ag\\\\\mu m_Ag = m_B g\\\\0.8(20)(9.8) = m_B (9.8)\\\\m_B = \frac{0.8(20)(9.8)}{9.8)} = \boxed{16 kg}$

The block must weigh <u>> 16 kg</u> for block A to move.

5 0

3 years ago

The number of oscillations that a wave completes per unit of time in called its _______.

Elena-2011 [213]

<span>The number of oscillations that a wave completes per unit of time is called "frequency".
Frequency represents in fact how many complete cycles of the wave are done in a single unit of time. It can be written also as
</span> $f= \frac{Ncycles}{time}$ <span>
</span>

6 0

3 years ago