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Tamiku [17]
3 years ago
8

A bookshelf is at rest in a room. a force of 35.0 newtons is applied to a bookshelf. if the floor imparts a frictional force of

2.90 newtons, what is the net force acting on the bookshelf?
Physics
2 answers:
Rasek [7]3 years ago
7 0
The net force (Fn) acting on the bookshelf is the difference between forces that are acting to it in opposite directions. In the given above, the frictional force, 2.90 N, always acts opposite to the force applied to it, 35 N.
 
                                       Fn = 35 N - 2.90 N = 32.1 N

Thus, the net force acting to the bookshelf is 32.1N. 



jeyben [28]3 years ago
7 0
The answer is 32.1 newtons
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5 0
3 years ago
In being served, a tennis ball is accelerated from rest to a speed of 42.1 m/s. The average power generated during the serve is
hoa [83]

Answer:

Force F = 69.35 N

Explanation:

given data

Ball Initial speed u = 0

Ball Final speed v = 42.1 m/s

average power generate = 2920 W

solution

Power generate is express as

P= \frac{W}{t}   ..............1

here W is work done and t is time

and work w = F × d

so

P= \frac{Fd}{t}  

and we know speed v = \frac{d}{t}  

so here

Power P = F × v

put here value and we get force

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8 0
4 years ago
This is a bond in which a single pair of electrons is shared between a pair of atoms.
Murrr4er [49]

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8 0
3 years ago
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

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        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

8 0
2 years ago
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