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Musya8 [376]
3 years ago
14

What is the momentum of a 31.2 kg object traveling at a velocity of 2.1 m/s?

Physics
1 answer:
harina [27]3 years ago
3 0

Answer:

p = 65.52 kg m/s

Explanation:

p = m × v

p = 31.2 × 2.1

p = 65.52 kg m/s

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Closing Summary Questions: Phases of Matter
snow_tiger [21]

Answer:

Explanation:

Particles in all states of matter are in constant motion and this is very rapid at room temperature. A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures.

Even with all of these state changes, it is important to remember that the substance stays the same—it is still water, which consists of two hydrogen atoms and one oxygen atom. Changing states of matter are only physical changes; the chemical properties of the matter stays the same regardless of its physical state!

5 0
3 years ago
En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)
Virty [35]

Answer:

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Explanation:

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7 0
3 years ago
How are primary and secondary succession similar and how are they different?
ExtremeBDS [4]
They are similar because they are all colors in the spectrum and they are different because you cant seperate primary colors but you can seperate secondary
8 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
3 years ago
An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e
kogti [31]
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
8 0
3 years ago
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