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il63 [147K]
3 years ago
15

A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a

llowed to come to rest, the vertical length of the shaik (rs53237) – work, energy, power – theisen – (20167) 2 spring is 3.62 m. calculate the elastic potential energy stored in the spring. answer in units of j
Physics
1 answer:
olganol [36]3 years ago
6 0
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
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Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant

F₁/x₁ = F₂/x₂
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Elastic work = 1/2 Fx
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3 years ago
When you close a switch in an electric circuit you are stopping the flow of electrons
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True because all switches use contacts to start or stop the flow of electrons in a circuit
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The boy moving down the hill posses what energy?​
elena-s [515]

Answer:    Kinetic energy

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Kinetic energy and potential energy can change forms. For example, the car moving up the hill is kinetic energy

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How fast is a 180 kg motorcycle traveling if it has 36000 J of kinetic energy??
leva [86]

Answer:

20 ms⁻¹

360 J

Explanation:

Kinetic energy is the energy possessed by a moving object solely due to its motion.

You can get the K.E. of an object using the equation,

K.E. =  (1/2)mv² where all terms in usual meaning

So you get,

K.E. = 36000 =  (1/2)×180×v²

v = 20 ms⁻¹

Also,

K.E. =  (1/2)×80×3² = 360 J

   

7 0
3 years ago
Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt
Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

= \frac{4}{3}\pi(6400+ 50)^3 -  \frac{4}{3}\pi (6400)&#10;^3\\\\=  \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3

Hence the volume of atmosphere is 2.6\times 10^{19} m^3

(b)

Write the ideal gas equation as foll ows:

PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles

no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules&#10;

Hence the required molecules is 13.3\times10^{43} molecules&#10;

(c)

Write the ideal gas equation as follows:

PV =nRT&#10;\\\\n=\frac{1.0 atm \times 0.5L&#10;}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles

no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}&#10;{Imole}= 1.2\times 10^{23} molecules

Hence the required molecules in Caesar breath is 1.2\times 10^{23} molecules

(d)

Volume fraction in Caesar last breath is as follows:  

Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}

(e)

Since the volume capacity of the human body is 500 mL.

Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath

5 0
3 years ago
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