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3 years ago

Which of the following is not an example of a cognitive process

Physics

2 answers:

baherus [9]3 years ago

5 0

The answer is D. Sleep

Tasya [4]3 years ago

4 0

Cognitive process means the process of thinking, and remembering so the awnser is, D. sleep.<span />

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You throw an orange out a widow at a height of 12 meters upwards at an angle of 32 to the horizontal with an initial velocity of

jeka94

Answer:

3 meters

Explanation:

add up dum boi

7 0

3 years ago

tensa zangetsu [6.8K]

Answer:

C. amount of charge on the source charge.

Explanation:

Electric field lines can be defined as a graphical representation of the vector field or electric field.

Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.

The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.

8 0

2 years ago

a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v

shutvik [7]

Isothermal Work = PVln(v₂/v₁)

PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = $e^{0.3098}$

v₂ = 19* $e^{0.3098}$

v₂ = 25.8999

v₂ ≈ 26 L Option b.

6 0

3 years ago

A soccer ball takes 20 s to roll 10 m. What is the average speed of the soccer ball.

Georgia [21]

Explanation:

velocity is distance divided by time.

the average speed of the ball is 10m/20s

= 0.5 m/s

8 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago