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ludmilkaskok [199]
2 years ago
10

A 3 kg rubber block is resting on wet concrete. The coefficient of static friction is 0.3. What is the minimum force that must b

e applied to the block to make the block begin to accelerate?
Physics
1 answer:
mrs_skeptik [129]2 years ago
8 0

Answer:

You would have to find the friction force of the rubber block which would be found with the equation of Normal force (mass*gravity) times cooeficient of friction which would give 8.82 N for the amount of friction and because you need more force than 8.82 N (assuming gravity is 9.8)

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Answer:

a) For y = 102 mA, R = 98.039 ohms

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According to ohm's law, V = IR

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b) y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}

y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]

y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right]

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

x = \left[\begin{array}{ccc}100\end{array}\right]

\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right] =  \left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] =  \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5}  \end{array}\right]

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