About how much of Earth's land surface is used for agriculture? 8% 18% 28% 38%

Being able to see a square in the middle of the image despite not having lines to form a square represents the Gestalt principle

vredina [299]

Answer: The correct answer for the blank is -

C. closure.

Gestalt principle of closure describes how we perceive complete figures even when the information that form the figure is missing.

This is due the fact that our brain responds to the familiar patterns inspite of getting incomplete information.

For instance, in the given question, we are able to perceive an image of square in the center despite not having actual lines that form a square.

Thus, it represents principle of closure.

7 0

3 years ago

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if abus travelling at 20m/s is subject to steady decceleration of 5m/s².how long will it take yo come to rest.

Likurg_2 [28]

Answer:

4 seconds

Explanation:

Given:

v₀ = 20 m/s

v = 0 m/s

a = -5 m/s²

Find: t

v = at + v₀

0 m/s = (-5 m/s²) t + 20 m/s

t = 4 s

6 0

3 years ago

when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why

Lady_Fox [76]

Answer:

hsvshxansjusjsnwjwisks

4 0

3 years ago

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While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?

mote1985 [20]

While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

(a) The speed of your phone after 4 s is:

V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

(b) The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

(c) If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

5 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

2 years ago