The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns

A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.

Answers

Answer 1

Answer:

C

Explanation:

Similarity


Related Questions

the rate of cooling determines ....... and ......​

Answers

Answer:

freezing point and melting point

A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.

Answers

Answer:

The resolving power remains same.

Explanation:

The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.

As the diameter is same but the focal length is doubled so the resolving power remains same.

a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point

Answers

Answer:

1.37 mm²

Explanation:

From the image attached below:

Let's take a look at the two rays r and r' hitting the same mirror from two different positions.

Let x be the distance between these rays.

[tex]d_o =[/tex] distance between object as well as the mirror

[tex]d_{eye}[/tex] = distance between mirror as well as the eye

Thus, the formula for determining the distance between these rays can be expressed as:

[tex]x = 2d_o tan \theta[/tex]

where; the distance between the eye of the observer and the image is:

[tex]s = d_o + d_{eye}[/tex]

Then, the tangent of the angle θ is:

[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]

replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:

[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]

[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]

[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]

x = (0.13157 × 10) mm

x = 1.32 mm

Finally, the area A = π r²

[tex]A = \pi(\frac{x}{2})^2[/tex]

[tex]A = \pi(\frac{1.32}{2})^2[/tex]

A = 1.37 mm²

A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.

Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.

Answers

The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.

Momentum is conserved, so the total momentum of the system is the same before and after the collision:

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

==>

(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'

==>

-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'

where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.

Kinetic energy is also conserved, so that

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' + 1/2 m₂ (v₂'

or

m₁ v₁² + m₂ v₂² = m₁ (v₁' + m₂ (v₂'

==>

(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' + (0.296 kg) (v₂'

==>

1.55 kg•m²/s² ≈ (0.160 kg) (v₁' + (0.296 kg) (v₂'

Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is

v₁' ≈ -3.11 m/s

v₂' ≈ -0.167 m/s

and take the absolute values to get the magnitudes.

If you want to instead use the masses from the "Required" section, you would end up with

v₁' ≈ -3.18 m/s

v₂' ≈ -0.236 m/s

Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day ​

Answers

what do you mean about it

Q)what are convex mirrors?​

Answers

Answer:

A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.

A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.

Determine the acceleration of a pendulum bob as it passes through an angle of 15 degrees to the right of the equilibrium point.

Answers

Answer:

Explanation:

Since energy is conserved:

2

mu  

2

 

=  

2

mv  

2

 

+mgh

⇒u  

2

=v  

2

+2gh

⇒(3)  

2

=v  

2

+2(9.8)(0.5−0.5cos60)

⇒v=2m/s

Acceleration of the simple pendulum is 2.62 m/s².

What is meant by a simple pendulum ?

When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.

Here,

Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.

Let T be the tension acting on the string.

As, the bob passes through the angle,

The weight of the bob becomes equal to the vertical component of the tension.

mg = T cos15°

Also, the horizontal component of the tension,

T sin15° = ma

By solving these two equations, we get that,

Acceleration of the simple pendulum,

a = g tan15°

a = 9.8 x 0.267

a = 2.62 m/s²

Hence,

Acceleration of the simple pendulum is 2.62 m/s².

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If a jet travels 350 m/s, how far will it travel each second?

Answers

Answer:

It will travel 350 meters each second.

Explanation:

The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.

Answer:

5.83 seconds

Explanation:

60 seconds in 1 minute

350 meters per second

350/60

=5.83

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.

Answers

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?

Answers

Answer:

2

Explanation:

pulling force because of it force

Answer:

5.9 cm

Explanation:

f: frequency of oscillation

frequency of oscillationk: spring constant

frequency of oscillationk: spring constantm: the mass

[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]

in this problem we know,

F= 1.4 Hz

m= 0.26 kg

By re-arranging the formula we get

[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]

The restoring force of the spring is:

F= kx

where

F= 1.2 N

k= 20.1 N/m

x: the displacement of the block

[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]

calculate the length of wire.

Answers

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

[tex]R = \frac{\rho L}{A}[/tex]

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]

L = 169.5 m

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.

Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?

Answers

Solution :

We know,

Distance,

[tex]$S=ut+\frac{1}{2}at^2$[/tex]

[tex]$S=ut+0.5(a)(t)^2$[/tex]

For the first 20 ms,

[tex]$S=0+0.5(220)(0.020)^2$[/tex]

S = 0.044 m

In the remaining 30 ms, it has constant velocity.

[tex]$v=u+at$[/tex]

[tex]$v=0+(220)(0.020)[/tex]

v = 4.4 m/s

Therefore,

[tex]$S=ut+0.5(a)(t)^2$[/tex]

[tex]$S'=4.4 \times 0.030[/tex]

S' = 0.132 m

So, the required distance is = S + S'

                                              = 0.044 + 0.132

                                              = 0.176 m

Therefore, the tongue can reach = 0.176 m or 17.6 cm

Answer:

The total distance is 0.176 m.

Explanation:

For t = 0 s to t = 20 ms

initial velocity, u = 0

acceleration, a = 220 m/s^2

time, t = 20 ms

Let the final speed is v.

Use first equation of motion

v = u + at

v = 0 + 220 x 0.02 = 4.4 m/s

Let the distance is s.

Use second equation of motion

[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]

Now the distance is

s' = v x t

s' = 4.4 x 0.03 = 0.132 m

The total distance is

S = s + s' = 0.044 + 0.132 = 0.176 m

Explain how blood circulation takes place in humans?​

Answers

Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.

Hope it helps you

Mark my answer as brainlist

have a nice day

Proper physical exercise makes bones _[blank 1]_.
People with stronger muscles and bones have better _[blank 2]_.

Which option shows the words that correctly fill in blank 1 and blank 2, in that order?


longer, flexibilitylonger, flexibility , ,

stronger, posturestronger, posture , ,

longer, posturelonger, posture , ,

stronger, flexibility

Answers

stronger, posturestronger, posture

hope that helped

. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?

Answers

Answer:

4

Explanation:

The cation that is reabsorbed from the urine in response to aldosterone

Answers

Answer:

If decreased blood pressure is detected, the adrenal gland is stimulated by these stretch receptors to release aldosterone, which increases sodium reabsorption from the urine, sweat, and the gut. This causes increased osmolarity in the extracellular fluid, which will eventually return blood pressure toward normal.

Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T

Answers

Answer:

 emf = 312 V

Explanation:

In this exercise the electromotive force is asked, for which we must use Faraday's law

           emf =  [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt

           Ф = B. A = B A cos θ

bold type indicates vectors.

They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values ​​1

It also indicates that the area is reduced from  a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear

            emf = -N B [tex]\frac{dA}{dT}[/tex]

            emf = - N B (A_f - A₀) / Dt

we calculate

           emf = - 60 1.60 (0 - 0.325) /0.100

           emf = 312 V

The direction of this voltage is exiting the page

Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.

Answers

Answer:

F = 1.128 10⁸ Pa

Explanation:

Pressure is defined by

         P = F / A

If the gas is ideal for equal force eds on all the walls, so on the piston area we have

        F = P A

We reduce the pressure to the SI system

       P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa

we calculate

       F = 150 10³ / 0.00133

       F = 1.128 10⁸ Pa

Why is it advised not to hold the thermometer by its bulb while reading it?

Answers

Because your body heat might change the temperature

Why we use semiconductor instead of metal in thermopile.

Answers

Answer:

Semiconductors are not normal materials. They have special properties which conductors/metals cannot exhibit. The main reason for the behavior of semiconductors is that they have paired charge carriers-the electron-hole pair. This is not available in metals.

the plane of a 5.0 cm by 8.0 cm rectangular loop wire is parallel to a 0.19 t magnetic field. if the loop carries a current of 6.2 amps, what is the magnitude of the torque on the loop

Answers

here is the state ment :0.2 loop

A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?

Answers

Answer:

The mass of the second block=0.457 kg

Explanation:

We are given that

m1=1.5 kg

v1=1.3m/s

v2=4.3 m/s

V=2.0 m/s

We have to find the mass of the second block.

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

Let m2=m

Substitute the values

[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]

[tex]1.95+4.3m=3+2m[/tex]

[tex]4.3m-2m=3-1.95[/tex]

[tex]2.3m=1.05[/tex]

[tex]m=\frac{1.05}{2.3}[/tex]

[tex]m=0.457 kg[/tex]

Hence,  the mass of the second block=0.457 kg

i don't understand this, can someone help please?? ​

Answers

Explanation:

N2 + H2 --> NH3

balance them:

N2 + 3 H2 --> 2 NH3

so if 6 moles of N2 react, 12 moles of NH3 will form.

(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )

Two forces are acting on a body. One acts east, the other at 35° north of east. If the
two forces are equal in magnitude of 50 N, find the resultant using the Law of Sines
and the Law of Cosines. Please answer with full solution. Thanks

Answers

A=B=50NAngle=theta=35°

We know

[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\Theta}}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{50^2+50^2+2(50)(50)cos35}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{2500+2500+2(2500)\times (-0.9)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000+5000(-0.9)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000+(-4500)}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{5000-4500}[/tex]

[tex]\\ \sf\longmapsto R=\sqrt{-500}[/tex]

[tex]\\ \sf\longmapsto R=22.4i[/tex]

Resultant using the Law of Sines and the Law of Cosines will be R=95 N

What is force?

Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.

The Magnitude of two forces =50 N

Angle between the forces = 35

By using the resultant formula

[tex]\rm R=\sqrt{A^2+B^2+2ABCos\theta}[/tex]

[tex]\rm R=\sqrt{50^2+50^2+2(50)(50)Cos35}[/tex]

[tex]\rm R=\sqrt{5000+5000(0.81)}[/tex]

[tex]\rm R=\sqrt{5000+4500}[/tex]

[tex]\rm R=95\ N[/tex]

Hence the Resultant using the Law of Sines and the Law of Cosines will be R=95 N

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S.I unit for distance =______

(A) m (B)cm

(c) km (d) mm

Answers

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

Answer:

A

Explanation:

A load of 25 kg is applied to the lower end and of a steal wire of length 25 m and thickness 3.0mm .The other end of wire is suspeded from a rigid support calculate strain and stress produced in the wire​

Answers

Answer:

the weight of the wire + 25kg

Explanation:

An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.

Answers

Answer:

Here, m=9×10

−31

kg,

q=1.6×10

−19

C,v=3×10

7

ms

−1

,

b=6×10

−4

T

r=

qB

mv

=

(1.6×10

−19

)(6×10

−4

)

(9×10

−31

)×(3×10

7

)

=0.28m

v=

2πr

v

=

2πm

Bq

=

2×(22/7)×9×10

−31

(6×10

−4

)×(1.6×10

−19

)

=1.7×10

7

Hz

Ek=

2

1

mv

2

=

2

1

×(9×10

−31

)×(3×10

7

)

2

J

=40.5×10

−17

J=

1.6×10

−16

40.5×10

−17

keV

=2.53keV

A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.

Select one:

a.
Space ship will have a velocity to the West and will be speeding up.

b.
Space ship will have a velocity to the East and will be speeding up.

c.
Space ship will have a velocity to the East and will be slowing down.

d.
Space ship will have a velocity to the West and will be slowing down.

e.
Ship experiences no change in motion.

Answers

Answer:

The correct answer is - c.  Spaceship will have a velocity to the East and will be slowing down.

Explanation:

In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.

As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.

5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?​

Answers

Answer:

gshshs

Explanation:

hshsksksksbsbbshd

Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point

Answers

Answer:

The right answer is "0.273 m".

Explanation:

Given:

Power (P),

[tex]\frac{1}{f} = 2D[/tex]

Near point,

u = 0.6 m

As we know,

⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]

By substituting the values, we get

⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]

            [tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]

            [tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]

            [tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]    

By applying cross-multiplication, we get

          [tex]0.6=2.2 \ v[/tex]

            [tex]v = \frac{0.6}{2.2}[/tex]

      [tex]S_{near} = 0.273 \ m[/tex]

Other Questions
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