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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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Explanation:

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3 years ago
Technician A says that the device that controls shift points on an automatic transmission is the valve body. Technician B says t
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The table represents the forces on four objects, with directions.
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Explanation:

An object is said to have balanced forces if the resultant of the forces acting on it is zero. This situation occurs, for instance, when the sum of the forces on the vertical direction is zero and the sum of the forces on the horizontal direction is zero as well.

For the objects in the table, we see that:

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Therefore, objects W and Y have balanced forces, and objects X and Z have unbalanced forces.

5 0
3 years ago
Read 2 more answers
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