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2 years ago

What is random motion? write any two examples

Physics

2 answers:

aleksley [76]2 years ago

7 0

Answer:

When an object in motion with no specific path and changes its motion suddenly is called random motion.

eg:Moving Clouds. Bees heading towards a flower in any direction. A rolling football during a game.

dalvyx [7]2 years ago

3 0

Answer:

<h3>Unpredictable kind of motion where an object moves in any direction and the direction keeps changing continuously.</h3>

<h2>example</h2><h2>---------</h2>

<h3>A rolling football during a game. </h3><h3>A bird movement in the sky. </h3><h3>The movement of fish in the sea.</h3><h3> </h3>

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A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new

icang [17]

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

$T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m$

Torque due to person

$T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m$

Now for total torque

$T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m$

4 0

3 years ago

A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from t

dalvyx [7]

Answer:

The net torque about the pivot is and the answer is 'c'

c. $T_{net}=8.58$

Explanation:

$T=F*d$

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

$T_1=F_1*d_1$

$T_1=7.8N*1.6m=12.48N*m$

$T_2=F_2*d_2$

$T_2=2.60N**cos(30)*3.0m$

$T_2= - 3.9 N*m$

Now to find the net Torque is the summation of both torques

$T_{net}=T_1+T_2$

$T_{net}=12.48N-3.9N=8.58N$

8 0

3 years ago

If a ball is dropped from a height of 10 m, hits the ground and bounces back up to a maximum height of 7.5 m, describe how the E

Greeley [361]

Answer:hmmmm

Explanation:

8 0

3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

3 years ago

What is random motion? write any two examples​

What is random motion? write any two examples