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3 years ago

10

Our ears receive sound waves and convert this energy into neural messages that travel to your brain and are processed as sounds.

This is an example of ________.
a
top-down processing
b
just noticeable difference
c
subliminal messaging
d
transduction

1 answer:

Radda [10]3 years ago

6 0

d

...........................

d

transduction

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250 mL

Explanation:

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Orbit circumference
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4 years ago

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the projectile will be ("E4" m

nordsb [41]

Answer:

$K=4\times 10^4\ J$

Explanation:

Given that,

Mass of cannon is,M = 2000 kg

Mass of projectile, m = 2 kg

Velocity of the projectile, v = 200 m/s

We need to find the kinetic energy of the projectile. We know that the formula for the kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2\\\\\text{Putting all the values, we get}\\\\K=\dfrac{1}{2}\times 2\times (200)^2\\\\K=40000\ J$

or

$K=4\times 10^4\ J$

So, the kinetic energy of the projectile will be $4\times 10^4\ J$ .

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3 years ago

If a toy car with a coin on top of it is rolling down a hill what force is keeping the coin on top of the car?

forsale [732]

The answer is gravity.

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3 years ago

If a rainstorm drops 5 cm of rain over an area of 13 km2 in the period of 3 hours, what is the momentum (in kg · m/s) of the rai

stiks02 [169]

To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.

The expression that defines the linear momentum is expressed as

P=mv

Where,

m=mass

v= velocity

According to our data we have to

v=10m/s

d=0.05m

$A=13*10^6m^2$

Volume $(V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3$

t = 3hours=10800s

$\rho = 1000kg/m^3$

From the given data we can calculate the volume of rain for 5 seconds

$V' = \frac{V}{t}*\Delta t_{total}$

Where,

$\Delta t_{total}$ It is the period of time we want to calculate total rainfall, that is

$V' = \frac{3.9*10^5}{10800}*5$

$V' = 1.805*10^2m^3$

Through water density we can now calculate the mass that fell during the 5 seconds:

$m' = V'*\rho$

$m' = 1.805*10^2*1000$

$m' = 1.805*10^5m^2$

Now applying the prevailing equation given we have to

$P=m'v$

$P = (1.805*10^5)(10)$

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Therefore the momentum of the rain that falls in five seconds is $1.805*10^6 Kg.m/s$

4 0

4 years ago