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Lesechka [4]
3 years ago
12

In the activation–synthesis model of dreaming, what is synthesized?

Physics
1 answer:
astra-53 [7]3 years ago
6 0

Answer:

hopefully that helps

Explanation:

B. a hodgepodge of memories and sensations that are internally generated by the brain

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A jet makes a landing traveling due east with a speed of 120 m/s .
vesna_86 [32]

Average acceleration over a time interval lasting \Delta t is

a_{\rm ave}=\dfrac{\Delta v}{\Delta t}

where \Delta v is the difference in the jet's final and initial velocities. It's coming to a rest, so

a_{\rm ave}=\dfrac{0-120\frac{\rm m}{\rm s}}{13.5\,\rm s}=-8.9\dfrac{\rm m}{\mathrm s^2}

so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).

7 0
4 years ago
A boy wants use his slingshot to shoot a water balloon into the air. His
Marrrta [24]

Answer:

A. 70 m OS the correct one

5 0
3 years ago
An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to
Lana71 [14]

complete question:

An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot

Answer:

a ≈  5281 ft

Explanation:

The observer at the top of a 462 ft cliff  measures the angle of depression from the top of the cliff to a point on the ground to be 5°.

The angle of depression form the top of the cliff = 5°

The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°.  90° - 5° = 85° Note sum of an angle on a right angle is 90°.  

using SOHCAHTOA  principle we can solve for the distance from the base of the cliff to the point on the ground(a)

tan  85° = opposite / adjacent

tan 85°  = a / 462

cross multiply

462 × tan 85° = a

a = 11.4300523 × 462

a =  5280.66  ft

a ≈  5281 ft

5 0
3 years ago
B. Projectile on cliff (range)
dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

         cos 25 = v₀ₓ / v₀

         sin 25 = Iv_{oy} / v₀

         v₀ₓ = v₀ cos 25

         v_{oy} = v₀ sin 25

         v₀ₓ = 22 cos 25 = 19.94 m / s

         v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

         y = y₀ + v_{oy} t - ½ g t²

     

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

         0 = 21 + 0.0192 t - ½ 9.81 t²

         4.905 t² - 0.0192 t - 21 = 0

         t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

        t = \frac{ 0.003914\  \pm \sqrt{0.003914^2 + 4 \ 4.2813 }   }{2}

        t = \frac{0.003914 \ \pm 4.13828}{2}

        t₁ = 2.07 s

        t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

        t = 2.07 s

now we can look up the distance traveled

         x = v₀ₓ t

         x = 19.94  2.07

         x = 41.28 m

5 0
3 years ago
b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept
olasank [31]

Answer:

fftgjjjuhgybcrgjjuhg ui

Explanation:

trggfyh

4 0
3 years ago
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