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sp2606 [1]
3 years ago
12

A book shelf is at rest in a room a force of 35.0 newton i applied to bookshelf

Physics
1 answer:
BigorU [14]3 years ago
4 0

The frictional force would be horizontal. If the 35.0 N force is also horizontal, those 2 forces would be in opposite directions. That would simplify the calculation. Since no mention is made of any angle, it is unlikely that the 35.0 N force is not horizontal.
Assuming that the 35.0 N force is horizontal, the net force is
35.0
N
−
2.90
N
=
32.1
N
.
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Ksivusya [100]

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5 0
3 years ago
The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris
Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

        \sum F_{x} = 0

        T + F_{AB} Sin 60 = 0 ...... (1)

       \sum F_{y} = 0

       F_{AB} Cos 60 + W = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

               \sigma_{allow} = \frac{T}{A}

       T = (20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}

          = 3925 kip

From equation (1),   F_{AB}Sin (60^{o})  = -3925

               F_{AB} \times -0.304 8 = -3925

             F_{AB} = 12877.29 kip

From equation (2),    -12877.29 (Cos 60) + W = 0

         -12877.29 kip \times \frac{1}{2} + W = 0

                           W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

5 0
3 years ago
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4 0
3 years ago
A 10 kg ball is held above a building with a height of 30 m. What is the
Darina [25.2K]

Answer: 2940 J

Explanation: solution attached:

PE= mgh

Substitute the values:

PE= 10kg x 9.8 m/s² x 30 m

= 2940 J

6 0
3 years ago
Which pair of equations can describe the path of a particle moving with an acceleration that is perpendicular to the velocity of
antoniya [11.8K]

Answer:

The question clearly describes the circular motion.

The circular motion equation is

a_{radial} = \frac{v^2}{r}

The path of the particle is circular.

Explanation:

In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.

6 0
3 years ago
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