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grin007 [14]
3 years ago
8

What is the velocity of position A and B?answer please ​

Physics
1 answer:
BaLLatris [955]3 years ago
5 0

Answer:

A = 15 m/s , B = 18.75 m/s

Explanation:

from the velocity is equal to zero ( at rest ) , you see that velocity is increasing by 3.75 m/s for each second.

I hope that it's a correct answer.

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Distinguish between directly proportional and inversely proportional with an example of each
RideAnS [48]

Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.

4 0
3 years ago
A sound wave of frequency 440 Hz is played by a musician. What is the
TiliK225 [7]

Answer:

343/440

Explanation:

Recall that v=d/t

Now, this is the same thing.

Frequency is 1/T and wavelength is the distance travelled in one period.

So Vs=f*λ

(the greek letter is used as the symbol of wavelength; it's arbitrary)

7 0
3 years ago
The number of wavelengths that pass a given point/second is called
myrzilka [38]

Answer:

I believe the answer is frequency

Hope this helps!! (let me know if it is right)

Explanation:

6 0
3 years ago
In class we described the tidal forces that are responsible for raising and lowering thewater level near the shore of the ocean.
Sloan [31]

The newton's law of universal gravitation used to describe how a particle attracts every other particle in the universe.

The equation is given by,

F= G\frac{m_1 m_2}{r_2}

Where,

F= Gravitational force

m_1, m_2 = masses of the objects

r= is the distance between the center of their masses

G= Gravitational constant

For our problem we have defined that,

M_m = 7.24*10^{22}kg (mass of the moon)

d= 3.84*10^8 km (distance Earth-moon)

G= 6.671*10^{-11}Nm^2/kg^2

M=Mass of a person

We have then,

F_m = M\frac{6.67*10^-11*7.34*10^22}{3.84*10^8}

F_m = M*3.320*10^{-5} N

In the other hand we have the force on m-mass due to earth

F_E= MG = M*9.8N

Ratio is given by

\frac{F_m}{F_E} = {M*3.320*10^{-5}}{9.8} = 3.387*10^{-6}

B) Suppose there is a group of young people surfing in the moonlight. They are directly under the moon. At this time the moon exerts its gravitational effect of the earth that causes the tide to rise. Around 6 hours later, when the earth has moved a quarter of the moon, the force on that point decreases, so the tide drops. However, after another 6 hours, people return and experience the same process. In this case the moon is not above them, but on the other side. This is because the moon having an orbit on the earth, generates an external force, similar to the previous one, but the earth reacts in the opposite way. It is like going in a car and turning it, all people will tend to get out of it, because a centrifugal force is experienced.

7 0
4 years ago
While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega
Paraphin [41]

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

 

   Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

   In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

   If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

4 0
3 years ago
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