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hjlf
3 years ago
5

Where is a water or a mechanical wave in which the particles in the medium move in a direction perpendicular to the direction of

energy transport identified another name for this type of wave motion
Physics
2 answers:
horrorfan [7]3 years ago
6 0

transverse waves. ripples on water surface

astra-53 [7]3 years ago
4 0

Answer:Transverse waves

Explanation: Its on USA TEST PREP

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An important aspect of fission reactions is that they produce _______________ which allow for chain reactions.
Alex777 [14]

An important aspect of fission reactions is that they produce free neutrons,  which causes chain reactions.

4 0
3 years ago
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The idea that earth was the center of the universe is an example of ___ that has long been discarded
aleksandrvk [35]
Hello. The answer to your question is ''<span>hypothesis''. I hope this helps! </span>
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3 years ago
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A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0
3 years ago
A star is in hydrostatic equilibrium when the outward push of pressure due to core burning is exactly in balance with the inward
viktelen [127]

Answer:

a and b

Explanation:

Hydro static equilibrium holds a star steady and balanced. Whenever a star stops burning hydrogen in its center, there must be evolutionary improvements to maintain equilibrium for the star Of example, if a star's internal pressure and temperature fall, gravity will take over and force the star to contract and heat up, restoring stability. By contrast, if a star's internal pressure and temperature rises, the extra pressure causes the star to widen and cool, restoring balance.

so, according to above explanation options a and b both are true

a) A small increase in the star's internal pressure and temperature causes the star's outer layers to expand and cool.

b) A small decrease in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.

7 0
3 years ago
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

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