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mario62 [17]
3 years ago
8

What are some examples of matter waves?

Physics
2 answers:
Elena-2011 [213]3 years ago
7 0

-- sound

-- ocean waves

-- seismic (earthquake) waves

-- waves in a rope

-- waves in a Slinky

-- waves in a guitar string

-- waves in the air inside a clarinet

zalisa [80]3 years ago
6 0

one example is sound waves.

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Only answer if you know all of them
lesya692 [45]
1)
The connections between neurons in the retina, specifically the connections referred to as “lateral inhibition,” help us see which of the following better?

<em><u>A) Contrast</u></em>
B) Faces
<span>C) Colors

2)
</span>Improving the contrast of an image (making the dark regions darker and the light regions lighter) helps us to identify:

<em><u>A) The edges of objects</u></em>
B) The center of objects
<span>C) The color of an object
</span>
3)
What assumption does our visual system make in order to see curved surfaces (domes, holes)?

<em><u>A) Light comes from above</u></em>
B) Curved surfaces are always evenly lit
<span>C) Curved surfaces are always easy to see, no assumptions are made
</span>
4)
Which part of the face does our brain pay the most attention to?

<u><em>A) Eyes and mouth</em></u>
B) Eyes and ears
<span>C) Eyes and chin
</span>
5)
If all these assumptions sometimes lead to mistakes, for example in these optical illusions, why do we make them?

A) It helps us see things faster
B) It helps us see things correctly
C) It helps us pay attention to what's important
<span><em><u>D) All of the above
</u></em>
</span>
Hope that helps :)
*the correct answers are bolded, italicized, and underlined.*
4 0
3 years ago
A 438 gram ball is traveling east at 39 m/s when it is hit by a 2.4 kg softball bat. After being in contact with the bat for 302
leonid [27]

Answer:

The magnitude of the impulse vector of the bat is 29.346 kg.m/s

The direction of the impulse vector of the bat is in the initial direction of the ball before impact

Explanation:

The given parameters are;

The mass of the ball, m₁ = 438 g

The speed of the ball = 39 m/s

The mass of the softball bat = 2.4 kg

The time of contact = 302 ms = 0.302 seconds

The speed of rebound = 28 m/s

The impulse = The change in momentum = Δp = F × Δt = m × Δv

The impulse on the ball = m₁ × Δv = 0.438 × (39 - (-28)) = 29.346 kg·m/s

Given that force of reaction of the bat is in opposite direction but equal to the force of the ball, and the time and the duration of contact with the ball is the same for both the ball and the bat, the impulse vector ore equal and opposite

Therefore, the magnitude of the impulse vector of the bat = 29.346 kg.m/s

The direction of the impulse vector of the bat = The initial direction of the ball before impact.

7 0
3 years ago
As the distance between two point charges is tripled, the electrostatic force between the charges will become
MatroZZZ [7]
Relation between electrostatic force and distance is inverse square i.e
     1
Fα ----
      r^2  

Hence if r is tripled, new electrostatic force will be 1/9 times old force.          
7 0
4 years ago
A major artery with a 1.7 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area
Alex

Answer:

0.14

Explanation:

Flow rate is the volume flowing through a point at a particular time, in calcuing flow rate we have

Q= v*t

it in terms of Area, we have Q= A*v

Where A= area

v= velocity.

Solving the question , flow rate is constant then

A*v= constant

A(i) v(i)= A(f) v(f)

Where A(i)= initial area= 1.00cm^2

A(f)= final area= 0.400cm^2

V(i) and V(f) are the initial and final velocity respectively and the ratio of the two will gives us the factor

Substitute the values into the equation we have

1 V(i)= 4 V(f)

But we were told that the cross sectional area of 1.00cm^2 branches into 18 smaller arteries.

Then

1 V(i)=0.4 V(f)*(18)

1 V(i)=7.2V(f)

Then if we find the ratio of the velocity, we will get the factor.

V(f)/V(i)= 1/7.2

V(f)/V(i)=0.14

Hence, the factor of the average velocity of the blood reduced when it passes into these branches is 0.14

8 0
3 years ago
at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

3 0
4 years ago
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