What is one result of meiosis?

A 1600-kg car travels at a speed of of 12.5 m/s. What is its kinetic energy?

Stella [2.4K]

KE= 0.5(m)(v^2)

KE= 250,000 J

6 0

3 years ago

In an extreme marathon, participants run a total of 100km; world-class athletes maintain a pace of 15 km/h. how many 230 Calorie

MAVERICK [17]

speed of the participants is given as

$v = 15 km/h$

$v = 15 \times \frac{1000 m}{3600s} = 4.2 m/s$

now the kinetic energy of the participant is given as

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}68\times 4.2^2$

$KE = 590.3 J$

As we know that

1 calorie = 4.2 J

$KE = 590.3\times \frac{1 cal}{4.2J} = 141.1 Cal$

so here 230 Calorie is provided by 1 bar

bar required is

$N = \frac{141.1}{230} = 0.61 bars$

7 0

3 years ago

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Melting Point of Substances

frez [133]

<span>(kg) Melting Point of Tin

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5 0

3 years ago

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Explain why an object’s weight is dependent upon where in the universe it is located.

strojnjashka [21]

Answer:

The weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

Explanation:

Everything obeys the law of universal gravity proposed by Newton. Where the attraction force with which one body attracts another depends on the mass of the body that attracts the body with less mass, the radius of the body with the greatest mass and the universal gravitational constant.

For a better understanding of this concept let us use examples with numeric values. In the first example we will determine with what force planet Earth attracts a person of 70 [kg].

And in the second example we will perform the same exercise but on a planet like Jupiter

Example 1:

A person with a mass of 70 [kg] is located on planet Earth which has a mass of 5.97x10^24 [kg] and a terrestrial radius of 6371 [km]. Find the force exerted by the planet upon the person.

We have the law of universal gravity

$F=G*\frac{m_{1} *m_{2} }{r^{2} } \\where\\G = 6.67*10^-11[\frac{N*m^{2} }{kg^{2}} ]\\m_{1}=70[kg]\\m_{2}=5.97*10^{24} [kg]\\r= 6371[km]$

Now replacing the values in the equation we have:

$F=6.67*10^{-11} *\frac{70*5.97*10^{24} }{(6371*10^3)^{2} } \\F=687[N]$

We can appreciate that if we only use the terms of mass of the earth, the gravitational constant and the radius of the earth, we will have the value of the gravity accelaration of the earth. Let's check

$g_{earth} =6.67*10^{-11} *\frac{5.97*10^{24} }{(6371*10^3)^{2} } \\g_{earth} = 9.81 [m/s^{2} ]$

Example 2:

A person with a mass of 70 [kg] is located on planet Jupiter which has a mass of 1.899x10^27 [kg] and a radius of 71492 [km]. Find the force exerted by the planet upon the person.

$F=6.67*10^{-11}*\frac{70*1.899*10^{27} }{(71492*10^3)^{2} } \\F= 1734.73[N]$

We will calculate the value of the gravity accelaration of Jupiter. Let's check

$g_{jupiter} = 6.67*10^{-11} *\frac{1.899*10^{27} }{(71492*10^3)^{2} } \\g_{jupiter}= 24.8 [m/s^2]\\$

Therefore an object in jupiter will weigh more than 2.5 times its weight than on planet Earth.

We found that the force of attraction or the weight of any object will depend on its location in the universe, commanded by the law of gravitacion universal de Newton

6 0

4 years ago

Which statement is true?

iogann1982 [59]

B
Think of inertia of getting into a car accident without a seat belt although the car stops you will not you would likely fly out the window

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3 years ago

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