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3 years ago

When sound waves cause the eardrum to vibrate, these vibrations are immediately passed on to the _______________.

Physics

2 answers:

dolphi86 [110]3 years ago

7 0

Answer: three small bones in the middle ear

Explanation: Sound waves enter the outer ear and travel through a narrow passageway called the ear canal, which leads to the eardrum. The eardrum vibrates from the incoming sound waves and sends these vibrations to three tiny bones in the middle ear. These bones are called the malleus, incus, and stapes.

Lilit [14]3 years ago

3 0

Answer:

Three small bones in the mid ear

Explanation:

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2 years ago

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What is the numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall?

Nuetrik [128]

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3 years ago

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A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

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3 years ago

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A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

CaHeK987 [17]

Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N

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3 years ago

A 250 g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room

zlopas [31]

Answer:

Q = 114895 J

Explanation:

To find the thermal energy gained by the ice you use the following formula:

$Q=mc(T_2-T_1)+H_f\ m$

m: mass of the ice = 0.250kg

T2: final temperature = 22°C

T1: initial temperature = -8.0°C

Hf: heat of fusion of water = 3.34*10^5 J/kg

c: specific heat of water = 4186 J/kg

By replacing the values of the parameters you have:

$Q=(0.250kg)(4186J/kg\°C)(22+8)\°C+(3.34*10^5 J/kg)(0.250kg)\\\\Q=114895\ J$

where you have considered that ice melts completely

3 0

3 years ago