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tankabanditka [31]
3 years ago
9

Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The c

oefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A)-0 B)-370JC) -190 J D-1800 J
Physics
2 answers:
m_a_m_a [10]3 years ago
6 0

Answer:

C) W = - 190 J

Explanation:

Notation

Wf = work done by the friction force (unknown)

Ff = force of the friction

d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m

Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

the network done by friction on a box that moves in a complete circle is 185.7 joules

Explanation:

Step one

Given

Radius of circle =1.82m

Circumference of the circle =2*pi*r

=2*3.142*1.82=11.43

Hence distance =11.43m

Coefficient of friction u=0.25

Weight of box =65N

We know that work =force*distance

But the limiting force =u*weight

Hence the net work done by friction

Wd=0.25*65*11.43

Wd=185.7 joules

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Five difference between elastic collision and inelastic collision?​
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Answer:

Elastic Collision

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2 years ago
A circuit contains four light bulbs. One light bulb goes out but the other three stays it. This must be an)
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Answer:

It is a parallel connection

Explanation:

In parallel connection the

Cell is not easily used up because the cells share the total current generated together with all bulbs.

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3 0
3 years ago
An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.
ivann1987 [24]
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
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</span>
5 0
3 years ago
One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-
choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = 0.12  \times 10^{-2} \ m

x = 1.17 \ cm = 1.17 \times 10^{-2}\ m

m = 149 kg

\delta = 7.87 \ g/cm^3

da = 2.28 \times 10^{-10}\ m

F_{net} = F-mg\\ \\0 = F - mg \\ \\  F = mg \\ \\ k_sx = mg \\ \\

∴

k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\  k_s = 124803.42  \ N /m

N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}

N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2

N_{chain} =  (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2

N_{chain} = 2.77 \times 10^{13}

N_{bond} = \dfrac{L}{da} \\ \\  = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}

\text{Finally; the stiffness of a single interatomic spring is:}

k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s

k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)

\mathbf{k_{si} =45.46 \ N/m}

4 0
3 years ago
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