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sergejj [24]
1 year ago
5

When you drop a object on earth, earth gets pulled slightly toward that object as well. How do I calculate the force if the eart

h being pulled?
Physics
1 answer:
kiruha [24]1 year ago
7 0

The gravitation force with which the earth is being pulled can be determined by applying Newton's law of universal gravitation.

<h3>What is gravitation force?</h3>

According Newton's law of universal gravitation, the force exerted between two objects in the universe is directly proportional to the product of masses of the two objects and inversely proportional to the square of the distance between the two objects.

Mathematically, the formula for gravitation force is given as;

F = GmM/R²

where;

  • m is the mass of the object
  • M is mass of earth
  • R is the distance of the object from earth
  • G is universal gravitation constant

If the mass of the object is know and the distance between earth  and the object is also known, the force with which the earth is being pulled can be calculated by applying Newton's law of universal gravitation as shown in the above equation.

Thus, the force with which the earth is being pulled can calculated as well.

Learn more about gravitation force here: brainly.com/question/27943482

#SPJ1

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Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

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