Physics Laws of Motion

The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is_________N.

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Contributor-Level 10

Kindly go through the solution

$\frac{F_{m a x}}{1 + 2} = 0.5 * 10 \Rightarrow F_{m a x} = 15 N$

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5 months ago

The initial mass of a rocket is 1000kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20ms^-². The gases come out at a relative speed of 500ms^-¹ with respect to the rocket: [Use g = 10m/s²]

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Vishal Baghel

Contributor-Level 10

$V_{r e l} \frac{d m}{d t} - m g = F_{e x t}$

=>500 $\frac{d m}{d t} - 1000 * 10 = 1000 * 20$

$\frac{d m}{d t} = 60 k g / s$

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5 months ago

If a book is at rest on a table, does Newton's First Law mean no forces are acting on it?

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Syed Aquib Ur Rahman

Contributor-Level 10

No, that's a common misconception about force in any object at rest. According to Newton's First Law, zero acceleration means zero net force. That does not mean zero force. A book on a table is at rest because two forces act on it. Gravity pulls it down. The table exerts an equal upward normal force. These forces balance each other, so the net force is zero. This can also be proved with Sir Isaac Newton's laws of motion. The 2nd Law implies that the book is not accelerating, so net force is zero. The third law of action and reaction states that the book pulls the table down with a force equal to its weight, as action. The table pushes

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A particle of mass m is suspended form a ceiling through a string of length L. The particle movies in a horizontal circle of radius r such that $r = \frac{L}{\sqrt[]{2}} .$ The speed of particle will be:

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Contributor-Level 10

$T s i n θ = \frac{m v^{2}}{r}$

T cosθ = mg

$t a n θ = \frac{v^{2}}{r g}$

$\Rightarrow v = \sqrt[]{r g}$

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5 months ago

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $ω .$ The coefficient of static friction between the bottom of the beaker and the surface of the disc is $ω .$ The beaker will revolve with the disc if :

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Raj Pandey

Contributor-Level 9

F. B. D of Beaker

By NLM2

$\begin{array}{l} f = m a = m ω^{2} R \\ \Rightarrow m ω^{2} R \leq μ N \\ R \leq μ g / ω^{2} \end{array}$

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5 months ago

If then the r $Z = \frac{A^{2} B^{3}}{C^{4}},$ elative error in Z will be:

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$\frac{Δ z}{z} = \frac{2 Δ A}{A} + \frac{3 Δ B}{B} + \frac{4 Δ C}{C}$

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5 months ago

A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is:

[Take g = 10m/s^-2]

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Contributor-Level 10

$a = μ g = 0.4 * 10 = 4 m / s^{2}$

Slipping stops when V_bag = V_conveyar

-> 0 + at = 2

-> t = $\frac{2}{a} = \frac{2}{4} = . 5 s e c$

So, x = $\frac{1}{2} a t^{2} = \frac{1}{2} * 4 * \frac{1}{4} = 0.5 m$

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5 months ago

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs^-1 on a conveyor belt moving with a velocity of 5ms^-1. The power needed to keep the belt moving with the same velocity will be:

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Contributor-Level 10

$P = \vec{F} . \vec{v} = v \frac{d P}{d t} = v^{2} (\frac{d m}{d t}) = 0.5 * 25 = 12.5 W$

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5 months ago

A block of mass 200g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F $= \sqrt[]{x}$ N as shown in figure.

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Contributor-Level 10

For equilibrium along the incline plane is given by,

F cos 60° = mg sin 60°

$F = \frac{0.2 * 10 * \sqrt[]{3}}{2 * 1} * 2$

$= 2 \sqrt[]{3} N = \sqrt[]{12} N$

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A curved in a level road has aradius 75m. The maximum speed of a car turning this curved road can be 30 m/s without skidding.; If radius of curved road is changed to 48m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be…….m/s.

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