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erastovalidia [21]
3 years ago
9

A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of ra

dius R. The small and large cars have accelerations as and aL respectively, as they travel around the curve. The magnitude of as is twice of that of aL. How does the speed of the small car VS compare to the speed of the large car VL as they move around the curve
Physics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

v_S=\sqrt{2}v_L

Explanation:

The acceleration experimented while taking a curve is the centripetal acceleration a=\frac{v^2}{r}. Since a_S=2a_L, we have that: \frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}

They take the same curve, so we have: r_S=r_L=R

Which means: v_S^2=2v_L^2

And finally we obtain: v_S=\sqrt{2}v_L

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Answer:

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8 0
3 years ago
Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each cha
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Answer:

<em>The force on each charge = 16 N</em>

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

<em>When d = 1 m, F = 1 N,</em>

<em>Constant: 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 2,</em>

<em>q₁q₂ = 1×1²/9×10⁹ </em>

<em>q₁q₂ = 1/9×10⁹  C²</em>

<em>When d = 1/4 m, q₁q₂ = 1/9×10⁹  C² and 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1</em>

<em>F =  9×10⁹×1/9×10⁹ /(1/4)²</em>

<em>F = 1/(1/16)</em>

<em>F = 16 N</em>

<em>Therefore the force on each charge = 16 N</em>

<em />

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3 years ago
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