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2 years ago

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Remy wonders if the height of the mountain has anything to do with the eventual size of the tsunami wave. How should Remy test t

his? A. Cause avalanches on several different mountains and see how large each resulting tsunami is. B. Build different-sized model mountains at the edge of a pool, pour sand down the side of each mountain, and see how large each resulting wave is. C. Pour different amounts of sand down the side of a model mountain into a pool and see how large each resulting wave is. D. Build a model town at the edge of a pool and see how large waves must be to cause damage to the buildings.

1 answer:

Anna71 [15]2 years ago

5 0

Answer:

<em>The answer is B</em>

Explanation:

<em>I got this from study island</em>

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A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

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${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

2 years ago

A test piolot flies with an acceleration of 5

Colt1911 [192]

On Earth, 1 g = 9.8 m/s² .

5 g = 5 · (9.8 m/s²)

5 g = 49 m/s²

5 0

2 years ago

Answer the following question

Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0

2 years ago

If you travel 1.7 km north from your house at noon, and at 6:00 PM you travel 5.4 km south, what is your displacement? 3.7 km no

Yakvenalex [24]

<u>Answer</u>

3.7 Km south

<u>Explanation</u>

The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.

Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).

Now lets add the two values.

(+1.7) + (-5.4) = 1.7 - 5.4

= - 3.7 Km But negative was towards south.

∴ Answer = 3.7 Km south.

6 0

2 years ago

A person jogs 4.0 km in 32 minutes, then 2.0 km in 22 minutes, and finally 1.0 km in 16 minutes. What is the jogger's average sp

kupik [55]

Answer:

22

Explanation:

i did it before

7 0

3 years ago