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Tom [10]
3 years ago
9

A 25-kilogram object is placed on a compression spring, and it creates a displacement of 0.15 meters. What is the weight of an o

bject that creates a displacement of 0.23 m on the same spring? Enter your answer as a number rounded to the nearest tenth, such as: 42.5

Physics
1 answer:
Effectus [21]3 years ago
4 0

Answer:

I hope it is correct ✌️

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The process of wind blowing sand from one location to another is called
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weathering

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The amount of diffraction depends on the size of the obstacle and the wavelength of the wave.
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John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds
katovenus [111]

Answer: ​Wernicke's aphasia

Explanation:

John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds it difficult to express his thoughts and cannot seem to find the right words to say while speaking. However, he can speak freely with proper syntax. In this scenario, John is most likely suffering from Wernicke's aphasia.

Wernicke's aphasia occurs when the leftward side of the middle of the brain is damaged or has been altered. An individual who suffers from Wernicke's aphasia will have difficulty in speaking in meaningful and coherent sentences or may have difficulty in understanding the speech of others.

6 0
3 years ago
A wave oscillates 5.0 times a second and has a speed of 4.0m/s what is the frequency of this wave
Viefleur [7K]

Answer:

The frequency of the wave is 5.0Hz

Explanation:

The frequency of a wave is the number of oscillations or revolutions made in a second.

7 0
3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
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