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Zolol [24]
3 years ago
5

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

friction is 0.340 N. How much work does your 1.55 N push do on the book
Physics
1 answer:
Rzqust [24]3 years ago
4 0

Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

                                       = 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

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A uranium and iron atom reside a distance R = 44.10 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
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Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

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iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron  distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × 10^{-19} C

and charge on iron will be q3 = 2 × 1.602 × 10^{-19} C

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and we know F = k\frac{q*q}{r^{2} }  

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Fu = Fi

k\frac{q*q}{r^{2} }  =  k\frac{q*q}{r^{2} }

put here k = 9*10^{9} and find r

9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }  =  9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }

\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}

r = 18.27 nm

distance r from the uranium atom is 18.27 nm

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