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Pavlova-9 [17]
3 years ago
5

Which of the following statements about the problem of object recognition is false?

Physics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

The answer is: letter c, in object recognition, the goal is recognizing the proximal stimulus.

Explanation:

Letter c is a "false" statement about object recognition because the goal is recognizing the distal stimulus and "not the proximal stimulus."

Distal stimulus refers to <em>an event or an object in the world that provides information to the proximal stimulus. </em>The proximal stimulus is a pattern of these events and objects that reaches to your senses. They can be registered in the person via<em> "sensory receptors." </em>

We need to recognize the distal stimulus and not the proximal stimulus. For example, when a lemon (distal stimulus) is being cut, it brings out a fragrance (proximal stimulus) that goes to the person's sense of smell. This gives the person a hint on where the smell is coming from and what it is. Then, the person recognizes that it is a lemon.

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A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as
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Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

T= 3600\ sec

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\dfrac{2\pi}{T}

Put the value into the formula

\omega=\dfrac{2\pi}{3600}

\omega=0.001745\ rad/s

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

a=r\omega^2

Put the value into the formula

a=0.55\times(0.001745)^2

a=1.675\times10^{-6}\ m/s^2

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

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3 years ago
A wave is produced from a vibrating string on a violin. Two observers are standing 20 m and 40 m away from the musician. The obs
max2010maxim [7]

Well the one that is closer can see and hear more

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3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

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