Ask question

Ask question

All categories

Alekssandra [29.7K]

3 years ago

6

The faintest sound an ear can hear (20 micro-pascals) is roughly ______ times smaller than atmospheric pressure at sea level.

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer: $5.06\times10^{9}$

At sea level, there is one standard atmospheric pressure which is equal to 101.325 kilopascals.

The pressure of faintest sound that a human ear can hear is 20 micro-pascals.

taking the ratio of two:

$\frac {101.325 kilopascals. }{20 micropascals} = \frac {101.325 \times 10^{3}Pa}{20 \times 10^{-6}Pa}$

$=5.06\times10^{9}$

Hence, the atmospheric pressure at sea level is $5.06\times10^{9}$ times greater than the faintest sound that a human ear can hear.

You might be interested in

Which of the following describes an action-reaction pair?

Diano4ka-milaya [45]

Answer B is the correct pair.

7 0

3 years ago

Why are renewable sources of energy better than nonrenewable sources?

Anton [14]

Because it does not produce waste, thus it doesn't harm the environment. also renewable sources are infinite.

5 0

3 years ago

Read 2 more answers

An object is placed in front of a convex mirror with a radius of curvature of magnitude 10 cm. The mirror produces an image that

jek_recluse [69]

Answer:

u = - 20 cm

m = $\frac{1}{5}$

Given:

Radius of curvature, R = 10 cm

image distance, v = 4 cm

Solution:

Focal length of the convex mirror, f:

f = $\frac{R}{2} = \frac{10}{2} = 5 cm$

Using Lens' maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Substitute the given values in the above formula:

$\frac{1}{5} = \frac{1}{u} + \frac{1}{4}$

$\frac{1}{u} = \frac{1}{5} - \frac{1}{4}$

u = - 20 cm

where

u = object distance

Now, magnification is the ratio of image distance to the object distance:

magnification, m = $\frac{|v|}{|u|}$

magnification, m = $\frac{|4|}{|-20|}$

m = $\frac{4}{20}$

m = $\frac{1}{5}$

4 0

3 years ago

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by θ = 0.220e3t, where

OLga [1]

Answer: $a_{t}=3.96$

$a_{c}=0.8712$

Explanation:

Given

$\theta =0.220e^{3t}$

r=2cm

Now angular velocity is given by $\omega =\frac{\mathrm{d}\theta}{\mathrm{d}t}$

$\omega =0.66e^{3t}$

Now linear velocity(v) is given = $\omega r$

$v=1.32e^{3t}$

Now tangential component of acceleration is given by

$a_{t}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=3.96e^{3t}$

at t=0

$a_{t}=3.96cm/s^2$

radial component of acceleration is given by

$a_{c}=\omega ^{2}r$

$a_{c}=0.4356e^{6t}\times 2$

$a_{c}=0.8712e^{6t} cm/s^{2}$

at t=0

$a_c=0.8712 cm/s^{2}$

5 0

3 years ago

Which diagram represents a closed circuit with the least resistance?

solniwko [45]

Answer:Circuit B is the answer

Explanation:

4 0

3 years ago

Read 2 more answers