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Setler [38]
3 years ago
5

The metric system is a measurement system that is: only used by a few of the nations of the world. used exclusively in chemical

calculations. commonly used by U.S. physical scientists. the official system for all nations of the world.
Physics
1 answer:
inysia [295]3 years ago
7 0

Answer:

The metric system is the official system for all nations of the world.

Explanation:

To make measures, it is necessary to have a set of standardized magnitudes to compare things we want to measure. This set of standardized magnitudes is called a unit system.

The metric system is a decimalized system of measurement which has meters and grams as a basic measurement unit. The metric system is the most used in the world, and also, is the official system for all nations of the world.

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Two tuning forks produce sounds of wavelengths of 3.4 meters and 3.3 meters.
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Explanation:

wavelength, λ = 3.4 m

wavelength, λ' = 3.3 m

Speed, v = 340 m/s

f = v / λ = 340 / 3.4 = 100 Hz

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2 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
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Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

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