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Arturiano [62]
3 years ago
12

In binocular rivalry, you see one image in the left eye and an incompatible image in the right eye. what do you perceive?

Physics
1 answer:
tankabanditka [31]3 years ago
8 0
What we perceive when we <span>see one image in the left eye and an incompatible image in the right eye using a binocular rivalry is that we usually see something halfway between one another. By definition, a binocular rivalry takes place when images gets alternated as presented in each eye.</span>
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Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini
kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

  • Mass of cart one M1  = 150kg
  • Initial Velocity U1 = 8m/s
  • Final VelocityV1 = 5 m/s

Mass of cart two M2  = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0
2 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
3 years ago
What is the main difference between magma and lava
Colt1911 [192]
Magma<span> is composed of molten rock and is stored in the Earth's crust. </span>Lava<span> is </span>magma<span> that reaches the surface of our planet through a volcano vent. </span>Lava<span>flow, Hawai.</span>
4 0
3 years ago
What is the current in a 160V circuit if the resistance is 2Ω? V= I= R =
sveta [45]

Answer:

80 amperes

Explanation:

Current in the circuit = ?

Voltage in the circuit = 160 Volts

Resistance = 2 Ω

Voltage = Current x Resistance

V = IR

160V = I x 2 Ω

I = 160V / 2 Ω

I = 80 Amperes

Therefore the current in the circuit is 80 amperes :)

8 0
3 years ago
A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
4 years ago
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