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3 years ago

9

A 50-N crate is pulled up a 5-m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to

the horizontal and there exists a constant frictional force of 10 N between the crate and the surface, what is the force applied by the worker?

1 answer:

rodikova [14]3 years ago

8 0

Answer:F=40.09 N

Explanation:

Given

weight of crate $W=50 N$

Inclination $\theta =37^{\circ}$

Frictional Force $f=10 N$

as the crate is moving with constant velocity therefore net Force on crate is zero

$F-50\sin (37)-f=0$

$F=50\sin (37)+10$

$F=30.09+10$

$F=40.09 N$

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Mass defect of each iron-56 nuclei:

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<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:

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<h3>b)</h3>

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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.

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

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

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

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