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Tju [1.3M]
3 years ago
5

A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with t

he ground of 35o. The coefficient of static friction between the box and the board is 0.5 and the coefficient of kinetic friction is 0.3 . A) Will the box slide down the ramp when it is at 35o? Prove it! B) How long will it take the box to slide to the other end of the ramp? (if it slides)
Physics
1 answer:
Vlada [557]3 years ago
3 0

Answer:

Explanation:

When the box is on the ramp , component of its weight along the ramp

= mg sinθ

Friction force acting on it in upward direction

=μ mg cosθ

For sliding

μ mg cosθ < mg sinθ

μ cosθ < sinθ

.5 x cos35 < sin35

.41 < .57

So the box will slide

When sliding starts , kinetic friction acts

Net force in downward direction

mgsinθ - μ mg cosθ

acceleration

= gsinθ - μ g cosθ

= 5.62 - .3 x 9.8 x cos35

= 5.62 - 2.4

= 3.22 m /s²

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Read 2 more answers
An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi
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Answer:

a

P_G  = 14.03 \  psig  

b

h_m =   0.148 \  m

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is P_{m} =  15.5 \  psig  =  15.5 *  6894.76 =  106868.78 \ N/m^2

The level of mercury is h  =  950 \ mm  =  0.950 \  m

The drop in the mercury level at the visible arm is d =  39.0 =  0.039 \  m

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

P_g  =  P_m  =  g *  \delta h  * \rho

Here \rho is the density of mercury with value \rho = 13.6 *10^{3} kg/m^3

and \delta h is the difference in the level of gas in arm one and two

So

\delta h  =  \frac{106868.78}{  13.6 *10^{3} *  9.8 }

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Generally from manometry principle we have that

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P_G  = \frac{ 9.6724 04 *10^{4} }{6894.76}

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