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denis-greek [22]

3 years ago

9

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.9 m/s2. For what value of the coe

fficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface

2 answers:

Luda [366]3 years ago

8 0

Answer:

The cabinet will slip along the bed surface for the values of coefficient of static friction lesser than 0.7

Explanation: Please see the attachments below

MaRussiya [10]3 years ago

3 0

Answer:

The coefficient of static friction is $\mu= 0.7667$

Explanation:

From the question we are told that

The acceleration is $a = 6.9 m/s^2$

The mathematical expression that show the relationship between frictional force and the sliding force of the cabinet at the point just before sliding begins is

$F_f = F_s$

Now $F_f = mg\mu$

Where m is the mass of the cabinet

g is the acceleration due to gravity

$\mu$ is the coefficient of static friction

And $F_s = ma$

Where a is the acceleration due to gravity

Now substituting this into the equation

$mg \mu = ma$

$g\mu =a$

$\mu = \frac{a}{g}$

Substituting values where $g =9.8m/s^2$

$\mu = \frac{6.9}{9.8}$

$\mu= 0.7667$

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Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the

Brrunno [24]

Answer:

$13.83928\times 10^{-6}\ F$

$249.10704\times 10^{-6}\ J$

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$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F$

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Total energy stored is $249.10704\times 10^{-6}\ J$

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$E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J$

Energy stored in C2 is $111.20842\times 10^{-6}\ J$

$E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J$

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$C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F$

$E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V$

The voltage would be 2.98273 V

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We need to check that before ladder slips the length of ladder the painter can climb.

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We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

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So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

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