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3 years ago

6

What is the resulting direction of a surface wave?

2 answers:

german3 years ago

7 0

surface wave is a wave that travels along the surface of a medium. The medium is the matter through which the wave travels. Ocean waves are the best-known examples of surface waves. They travel on the surface of the water between the ocean and the air.

HOPE IT HELPS

sweet-ann [11.9K]3 years ago

5 0

This question is incomplete. Here is the complete question:

What is the resulting direction of a surface wave

A: Opposite

B: Circular

C: Parallel

D: Perpendicular

Answer:

The correct answer is option B. Circular.

Explanation:

This question can sometimes be confusing. The fact is that the surface waves go perpendicular and parallel as well, which results in their direction becoming CIRCULAR.

These types of waves usually appear when an earthquake occurs near the Earth's surface, and they move on this surface.

One of the types of surface waves is ocean waves. These waves move in a circular way.

To see how their movement is, you can search an internet search engine for a gif of “surface waves” and you can better understand this explanation.

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Which data set has the largest range?

MrRissso [65]

Answer:

Option C

Explanation:

We have to check range of all options first

For A:

Largest Value: 5

Smallest Value: 1

So range = Largest value - smallest value

5-1 = 4

For B:

Largest Value: 6

Smallest Value: 4

Range = 6-4 = 2

For C:

Largest Value: 9

Smallest Value: 1

Range = 9-1 = 8

For D:

Largest Value = 9

Smallest Value = 3

Range = 9-3=6

So, the data set in option C has the largest range

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2 years ago

Can the velocity of a body revese the direction when acceleration is constant?

TEA [102]

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

7 0

2 years ago

Help pls lol and tysm

Zanzabum

Answer:

umm section 2 and 4

Explanation:

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2 years ago

Anyone know what the answer is?..

Licemer1 [7]

Answer:

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3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

2 years ago