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3 years ago

10

Without this part of our brain our short term memories would never become engrams a) Thalamus b) Hypothalamus c) Hippocampus d)

Corpus Callosum

1 answer:

Marat540 [252]3 years ago

5 0

Corpus Callosum is related with short term memory,

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Explain why the sound produced by every vibrating body cannot be heard by us ?

VMariaS [17]

Answer:

The human ear is not sensitive to every frequency of sound, rather, our hearing range is from 20Hz to 20,000Hz. This means sound frequencies outside this range are not audible to the human ear, which is why not every sound produced is heard. This is quite a blessing because if we could hear every vibrating body, then our ears would never stop listening: we won't be able to distinguish 'important sounds' from unimportant ones, rendering our ears effectively useless. Perhaps this is why we have a hearing range in the first place: sounds of our friends and our foes lie in 20Hz-20kHz range.

5 0

4 years ago

adelina 88 [10]

Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

1.1m to 0.04125 m

Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

wavelength = 330/300=1.1 m

wavelength = 330/8000 = 0.04125m

note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

6 0

3 years ago

You may make the following measurements of an object 42kg and 22m3. What would the objects density be?

ki77a [65]

Density=mass/volume
to find the density
mass=42kg
volume=22m3
so density=42/22
density=1.9Kgm3

8 0

4 years ago

In a ________________ wave, such as a sound wave, the particles in the medium vibrate in the same direction that the wave travel

Varvara68 [4.7K]

The answer would be A, transverse.

6 0

3 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago