Un auto se mueve en una trayectoria circular de radio 45,7 m,como observamos en la figura. Si el coeficiente de rozamientoestático entre la rueda del auto y la carretera es μₑ ¿ quevelocidad magnitud en magnitud debe llevar el auto para nopatinar?.

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Ответ:

sophiaa23

05.09.2022

Physics

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= 0.03i+0.05j

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q= $1.6\times 10^{-19} C$

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

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