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NeTakaya
3 years ago
12

Why is an element considered a pure substance?

Physics
1 answer:
anastassius [24]3 years ago
3 0

Answer:

An element is a pure substance and is made of only one type of atom.

Explanation:

It cannot be broken down into a simpler substance.

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<span>Skipping OTF and considering a small angle, then 2.38E5 • 1.22(4.56E-7/2.2E-3): 60 meters.</span>
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Which could describe the motion of an object?
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What factors that affect magnetic force
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Strong electrical currents in close proximity to the magnet.

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Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The on
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Answer: (a)F=7(10)^{-7}N

              (b)F=1.344(10)^{-6}N  

              (c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}   (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father m_{1}=100kg, the mass of the baby m_{2}=4.20kg and the distance between them r=0.2m, the force F_{F}  exerted by the father is:

F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}   (2)

F_{F}=0.0000007N=7(10)^{-7}N   (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter m_{J}=1.898(10)^{27}kg, the mass of the baby m_{2}=4.20kg and the distance between Jupiter and Earth (where the baby is) r_{E}=6.29(10)^{11}m, the force F_{J}  exerted by the Jupiter is:

F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}   (4)

F_{J}=0.000001344N=1.344(10)^{-6}N   (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

F_{J}=0.000001344N=1.344(10)^{-6}N   and F_{F}=0.0000007N=7(10)^{-7}N  we can see F_{J} is greater than F_{F}. However, the difference is quite small as well as the force exerted on the baby.

7 0
3 years ago
Three moles of an ideal gas are taken around the eyele acb shown in Fig.For this gas Co-29.2J/mol-K.Process ac is at constant pr
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For three moles of an ideal gas are taken around the eye, the total work for the cycle is mathematically given as

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<h3>What is the total work F for the cycle.(R=8.31J/molK)?</h3>

Generally, the equation for work  is mathematically given as

Wt=wac+Wc+Wba

Therefore

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In conclusion,

Wt=wac+Wc+Wba

Wt=4788.864J+0+(-6734.34J)

Wt=+1945.475J

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