All categories

2 years ago

If you ride quickly down a hill on a bicycle your eardrums are pushed in before they pop back. Why is this?

Physics

1 answer:

Mice21 [21]2 years ago

7 0

Answer:

<em>The difference in pressure between the external air pressure, and the internal air pressure of the middle ear.</em>

Explanation:

First of all, we should note that pressure decreases with height and increases with depth. The air within the middle ear (between the ear drum and the Eustachian tube) adjusts itself to respond to the atmospheric pressure, or when we yawn. At a high altitude like on the hill, the air pressure in the middle ear, is fairly low (this is to balance the low air pressure at this height). While riding down the hill quickly, there is little time for the air pressure in the ear to readjust itself to the increasing external air pressure, causing the external air to push into the ear drum. Along the way, the air within the middle ear is adjusted by the opening of the Eustachian tube, allowing more air into the space in the middle ear to balance the external air pressure. This readjustment causes the ear to pop.

You might be interested in

A charge Q is transferred from an initially uncharged plastic ball to an identical ball 10.0 cm away. The force of attraction is

victus00 [196]

Answer:

a charge Q is transferred from an initially uncharged

Explanation:

Hope this helps!

5 0

2 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

8 0

2 years ago

A steel rod is pulled in tension with a stress that is less than the yield strength. The modulus of elasticity may be calculated

pshichka [43]

Answer:

B. Axial stress divided by axial strain

Explanation:

Elasticity:

It is the tendency of an object to deform along the axis when an opposing force is applied without facing permanent change in shape.

Plasticity:

When an object crosses the elasticity limit, it enters plasticity where the change due to stress is permanent and the object might even break.

Yield strength:

Yield strength is the point of maximum bearable stress that indicates the limit of elasticity.

Our case:

As the stress applied is less than the yield strength, the rod is still in the elasticity state and its modulus can be calculated.

Modulus of Elasticity = Stress along axis/Ratio of change in length to original length

Axial strain is basically the ratio of change in length to original length.

So, Modulus of Elasticity = Axial Stress/ Axial Strain

6 0

3 years ago

A cannon can make a maximum angle of 30 degrees with the horizon. What is the minimum speed of a cannon ball if it must clear a

stiks02 [169]

Answer:

28.3 m/s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 30°

Maximum height (H) = 10 m

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) =?

Thus, we can obtain the minimum velocity cannon ball by using the following formula:

H = u²Sine² θ / 2g

10 = u² × (Sine 30)² / 2× 10

10 = u² × (0.5)² / 20

10 = u² × 0.25 / 20

10 = u² × 0.0125

Divide both side by 0.0125

u² = 10/ 0.0125

u² = 800

Take the square root of both side

u = √800

u = 28.3 m/s

Therefore, the minimum speed of the cannon ball is 28.3 m/s

8 0

2 years ago

A soccer ball is rolling with a constant acceleration of -1.5 m/s2 . At ti = 0 s, the ball has an instantaneous velocity of 14 m

miskamm [114]

Answer:

$v_{f} = 9.5\,\frac{m}{s}$

Explanation:

The final velocity is:

$v_{f} = v_{o} + a \cdot t$

$v_{f} = 14\,\frac{m}{s} + (-1.5\,\frac{m}{s^{2}} )\cdot (3\,s)$

$v_{f} = 9.5\,\frac{m}{s}$

6 0

2 years ago