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blagie [28]
2 years ago
14

A very smart 3-year-old child is given a wagon for her birthday. She refuses to use it. "After all," she says, "Newton's third l

aw says that no matter how hard I pull, the wagon will exert an equal but opposite force on me. So I will never be able to get it to move forward." What would you say to her in reply?
Physics
1 answer:
natita [175]2 years ago
5 0

Answer:

Explanation:

She's correct but doesn't mean the wagon cannot put into motion. The force that she applied on the wagon, according to Newton's 2nd law, would have generated an acceleration, which translates into motion. The reaction force the wagon applies on her due to Newton's 3rd law, would not hinder its own motion.

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Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.
user100 [1]
Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.
4 0
2 years ago
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Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u>  m.

When the speed of the bottle is 3 m/s, the average maximum height of the beanbag is<u> 0.43</u>  m.

When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is  <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is  <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is  <u>1.86</u> m.

Sorry for not answering early on! If anyone in the future needs help, I got these answers from 2020 egenuity, though I can't post the picture for proof. Stay Safe!

6 0
2 years ago
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the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce
Tanzania [10]

Answer:

The minutes hand travels 39.60 cm.

Explanation:

Note: a clock has a shape of a circle, the minutes hand is the radius, and the travel of the minutes hand forms a arc.

Length of an arc = ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Where L = length of an arc, ∅ = angle formed by an arc, r = radius of the arc.

Given:  ∅ = 252°, r = 9 cm, π = 3.143.

Substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus the minutes hand travels 39.60 cm.

4 0
2 years ago
A 12 volt car battery produces more current than a 1.5 volt battery for a TV remote because...
gavmur [86]

This question is loaded with misleading, marginally meaningful phrases.

A 12V battery produces more current than a 1.5V battery produces,
through the same device, because the potential difference between
the terminals of the 12V battery is 8 times the potential difference
between the terminals of the 1.5V battery.

Of the choices offered, choice 'C' is the closest.  But it should say
"electrical potential", NOT "potential energy".
 
3 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
2 years ago
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