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svp [43]
3 years ago
14

As a box is pushed 30 meters across a horizontal floor by a constant horizontal force of 25 newtons, the kinetic energy of the b

ox increases by 300 joules.
How much total internal energy is produced during this process?
Physics
1 answer:
irakobra [83]3 years ago
5 0

Answer:

1,050 Joules

Explanation:

<u>Step 1:</u> work done in moving the box 30 meters

work done = force X distance

                  = 25N X 30 = 750 Joules

<u>Step 2: </u>calculate total internal energy

Total internal energy = work done + kinetic energy

                                   = 750 Joules + 300 Joules

                                   = 1,050 Joules = 1.05 KJ

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Explanation:

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3 years ago
An 80-kg clown sits on a 20-kg bike on a tightrope attached between two trees. The center of mass of the clown is 1.6 m above th
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Answer:

Explanation:

the center of mass formula

Ycm= [(m₁y₁) + (m₂y₂) + (m₃y₃)] / (m₁+m₂+m₃)

Rope forms the x axis and position of centre of different massses are above or below it so they represent their location on y - axis.

y₁ = 1.6 , y₂ = .7 and y₃ = - 2.1

Ycm ( given ) = - .5

Putting the values of masses and positions

- .5 = 80 x 1.6 + 20 x .7 + m₃ x - 2.1 / ( 80 + 20 + m₃ )

- .5 = 128  + 14  + m₃ x - 2.1 / ( 100+ m₃ )

- 50 - .5 m₃ = 142 - 2.1 m₃

1.6 m₃ = 192

m₃ = 120 kg .

B )

Total downward force is weight of  total mass  = 80 + 20 + 120

= 220 kg

weight = 220  x 9.8 = 2156 N .

component of weight perpendicular to rope

= 2156 cos 15 = 2082.53 N

This force will be equally distributed over each tree , so force on each tree =  2082.53 / 2 = 1041.26 N .

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3 years ago
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the process of making alloys involves _____ pure metals to remove impurities. Then the pure metals are ____ with other component
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5 0
3 years ago
Read 2 more answers
Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u
SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

  • Proton rest mass: \rm 1.0072765\;amu;
  • Neutron rest mass: \rm 1.0086649\; amu.
  • Speed of light in vacuum: \rm 2.99792458\times 10^{8}\;m\cdot s^{-1}.
  • Charge on an electron: \rm 1.6021765\times 10^{-19}\;C.

<h3>a)</h3>

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other 56 - 26 = 30 particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu.

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

\rm 56.464736 - 55.934939 = 0.514197\;amu.

<h3>b)</h3>

By the mass-energy equivalence,

E = m\cdot c^{2}.

Refer to this equation, the speed of light in vacuum c^{2} is the conversion factor between mass m and energy E. The value of c is usually given only in SI units \rm m\cdot s^{-1}. Accordingly, the value of c^{2} will be in the SI unit \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}.

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.  

\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}.

Convert the unit of c^{2} from \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1} to the desired \rm MeV \cdot amu^{-1}:

\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}.

Total binding energy in each iron-56 nucleus:

\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}.

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 \mathrm{^{56}Fe} will be:

\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV.

6 0
3 years ago
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