Chapter 3: Problem 377
A car turns a corner on a slippery road at a constant speed of $10 \mathrm{~m} / \mathrm{s}\(. If the coefficient of friction is \)0.5$, the minimum radius of the arc at which the car turns is meter. (A) 20 (B) 10 (C) 5 (D) 4
Chapter 3: Problem 377
A car turns a corner on a slippery road at a constant speed of $10 \mathrm{~m} / \mathrm{s}\(. If the coefficient of friction is \)0.5$, the minimum radius of the arc at which the car turns is meter. (A) 20 (B) 10 (C) 5 (D) 4
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Get started for freeMatch the column \begin{tabular}{l|l} Column - I & Column - II \end{tabular} (a) Body line on a (p) is a self adjusting horizontal surface \(\quad\) force (b) Static friction (q) is a maximum value of static friction (c) Limiting friction (r) is then limiting friction (d) Dynamic friction (s) force of friction \(=0\) (A) $\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{p}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{r}$ (B) $\mathrm{a}-\mathrm{p}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{r}, \mathrm{d}-\mathrm{s}$ (C) $\mathrm{a}-\mathrm{s}, \mathrm{b}-\mathrm{r}, \mathrm{c}-\mathrm{q}, \mathrm{d}-\mathrm{p}$ (D) $\mathrm{a}-\mathrm{r}, \mathrm{b}-\mathrm{q}, \mathrm{c}-\mathrm{p}, \mathrm{d}-\mathrm{s}$
The force acting on a body whose linear momentum changes by $20 \mathrm{kgms}^{-1}\( in \)10 \mathrm{sec}$ is (A) \(2 \mathrm{~N}\) (B) \(20 \mathrm{~N}\) (C) \(200 \mathrm{~N}\) (D) \(0.2 \mathrm{~N}\)
A lift of mass \(1000 \mathrm{~kg}\) is moving with an acceleration of $1 \mathrm{~ms}^{-2}$ in upward direction Tension developed in the rope of lift is \(\mathrm{N}\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)\) (A) 9800 (B) 10,000 (C) 10,800 (D) 11,000
The motion of a particle of a mass \(m\) is describe by \(\mathrm{y}=\mathrm{ut}+(1 / 2) \mathrm{gt}^{2}\). Find the force acting on the particle. (A) \(\mathrm{F}=\mathrm{ma}\) (B) \(\mathrm{F}=\mathrm{mg} \quad\) (C) \(\mathrm{F}=0\) (D) None of these
A body of mass \(5 \mathrm{~kg}\) starts from the origin with an initial velocity \(u^{\rightarrow}=30 \mathrm{i}+40 \mathrm{j} \mathrm{ms}^{-1}\). If a constant Force $\underline{F}=-\left(\mathrm{i}^{\wedge}+5 \mathrm{j}\right) \mathrm{N}$ acts on the body, the time in which the y-component of the velocity becomes zero is (A) \(5 \mathrm{~s}\) (B) \(20 \mathrm{~s}\) (C) \(40 \mathrm{~s}\) (D) \(80 \mathrm{~s}\)
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