All categories

3 years ago

A uniform brick of length 25 m is placed over

Physics

1 answer:

Harlamova29_29 [7]3 years ago

3 0

The system of two rods will lie on the table as show in the figure

The center of first rod will lie exactly at the edge of first rod and then the center of mass of two combined rods will lie at the edge of the table.

So now the whole system will rest on the rod and it will not tipping off.

Since both rods are identical so we can say that the system will have its center of mass at the mid point on the line joining the two centers

So the value of x will be at mid point of line joining the two points on rod

$x = \frac{12.5}{2} = 6.25 cm$

So total length over the edge will be given as

$L = 6.25 cm + 12.5 cm = 18.75 cm$

You might be interested in

sticking your fingers into a wall socket will not bring you into direct contact with an electrical current. true or false

SCORPION-xisa [38]

The correct answer is false cause how can u fit your finger in a wall something it's to small

3 0

3 years ago

Read 2 more answers

Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus

poizon [28]

Answer:

For destructive interference phase difference is

$(2n+1)\pi$ where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of $\frac{\pi}{4}$ would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of $( \frac{\pi}{4} + 2\pi)$ because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of $\pi$ or which is same as $(2\pi+\pi)$

Generalizing to:

a phase difference of $(2n+1)\pi$ where n∈ {W}

{W}= set of whole numbers.

3 0

3 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows: