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faust18 [17]
3 years ago
7

We have seen that when you generate a wave on a stretched spring, as long as the medium doesn’t change (that is, the tension and

the linear mass density remain the same), the speed of the wave on the spring is constant. Does this mean that the particles of the spring always have zero acceleration? Explain why or why not.
Physics
1 answer:
TiliK225 [7]3 years ago
6 0

Answer:

No, the acceleration is not always zero.

Explanation:

It does not mean that the acceleration of the particle is zero.

The velocity of wave is different from the velocity of particle.

The acceleration of wave is different from the acceleration of particle.

the acceleration of the particle is given by

a =- w^2 y

where, w is the angular frequency and y is the displacement from the mean position.

So, the acceleration is zero at mean position only and it varies as the position changes.  

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An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. I
Mashutka [201]

Answer:

the water level remains same

Explanation:

This can be explained by Archimedes's principle which says that the wood will sink if weight of wood is more than the weight of the water displaced with weight equal to the water displaced otherwise the wood will float.

Therefore, buoyancy or the buoyant force is the same as the weight of wood, the weight of the water displaced by wood is also the same as that of the weight of wood.

Thus, we can see that the weight of the wood remains same and so is the level of water.

6 0
3 years ago
How to find id in this app. ​
AysviL [449]

Go and click to the invitation bar and you can find an option written as " search friends " . Then it's easy to find that unknown user if you're pretty fond with his/her username and DP ( display picture ).

6 0
3 years ago
Can someone please help me with this question? ASAP thank you!
Assoli18 [71]

did you get the answer

8 0
3 years ago
A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part
mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

4 0
2 years ago
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