when we touch a steel rod and a paper simultaneously we feel that the rod is comparatively colder .why?

A 0.150 kg ball on the end of a 1.10 m long cord (negligible mass)is swung in a vertical circle..

Aneli [31]

<span> For any body to move in a circle it requires the centripetal force (mv^2)/r. In this case a ball is moving in a vertical circle swung by a mass less cord. At the top of its arc if we draw its free body diagram and equate the forces in radial direction to the centripetal force we get it as T +mg =(mv^2)/r T is tension in cord m is mass of ball r is length of cord (radius of the vertical circle) To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s In the second case the speed of ball at top = (2*3.285) =6.57 m/s Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom we get velocity at bottom as 9.3m/s. Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force T-mg=(mv^2)/r We get tension in cord T=13.27 N</span>

3 0

3 years ago

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A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

Cellular phones use _____?

dangina [55]

Let's see: frequency of cellular phone waves (GSM phones) is (800-1900 MHz). If we look at the table of the electromagnetic spectrum, we can see that this range is contained within the frequencies of the microwaves, which include waves in the range 300 MHz-300 GHz.

So, summarizing, the correct answer is "microwaves".

3 0

3 years ago

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What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7

Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration $n= 2.7\times10^{18}\ cm^{-3}$