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sveticcg [70]
3 years ago
13

What isFigure ground in psychology

Physics
1 answer:
vivado [14]3 years ago
3 0

Answer:

Figure-ground organization is a tupe of perceptual grouping that is a vital necessity for recognizing objects trough vision.

You might be interested in
if a man has a mass of 83 kilograms on earth, what will the force of gravity on his body be on the moon
AnnZ [28]

If a man has a mass of 83 kilograms on Earth, the force of gravity on his body be on the moon 135.6N. force =mass*acc , 83 * 9.8/6= 813.4/6 = 135.6N

3 0
3 years ago
Read 2 more answers
Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl
Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of  acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = (\frac{(m₂-m₁)}({m₁+m₂} * g = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

6 0
3 years ago
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
What is the frequency of light with a wavelength of 3 meters?
blsea [12.9K]

Answer:

3x10^8=3f f=1x10^8 It think it is hopeful

7 0
3 years ago
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
miv72 [106K]

Answer:

C.   In the first collision has twice the momentum as when it stays still ( second colllions)

Explanation:

To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment

     I = Δp = m vf - m v₀

     I = m (vf -v₀)

Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)

    I₁ = m (-0.3 - 0.3)

    I₁ = -0.6 m

Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)

    I₂ = m (0 - 0.3)

    I₂ = -0.3 m

Let's calculate the relationship between the two impulses

     I₁ / I₂ = -0.6m / -0.3m

     I₂ / I₂ = 2

When it bounces it has twice the momentum as when it stays still

Now let's analyze the answers:

A.   False The momentum changes

B. False. The momentum is less in the second collision

C. True.  The momentum is double in this collision

D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations

E. False.  When they say bounces it implies the same speed with the opposite direction

4 0
3 years ago
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