What element is used as a lining in aprons to protect people in x rays

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Vlada [557]

Answer:

$Ep=-2.3*10^{-19}J$

Explanation:

The change in potential energy can be expressed as:

$Ep=K.\frac{q1.q2}{r}$

where K is a constant with a value of $9*10^{9}\frac{N.m^{2}}{C^{2}}$ , q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is $+1.6*10^{-19}C$ and the charge for the electron is $-1.6*10^{-19}C$ .

Converting r=1.0nm to m:

$1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m$

Replacing values:

$Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}$

$Ep=-2.3*10^{-19}J$

5 0

2 years ago

If a ball rolls at constant velocity, it is rolling at constant speed? explain

Shtirlitz [24]

Velocity is speed with direction. So, if velocity varies directly with speed, that statement would be true. A constant velocity would resort in a constant speed. They are connected and are dependant on each other.

I hope this helps!
~kaikers

7 0

3 years ago

2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the

ASHA 777 [7]

The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.

i) The angular acceleration of the disc ( $\alpha$ ), in revolutions per square second, is found by the following kinematic formula:

$\alpha = \frac{\omega_{f}-\omega_{o}}{t}$ (1)

Where:

$\omega_{o}$ - Initial angular speed, in revolutions per second.
$\omega_{f}$ - Final angular speed, in revolutions per second.
$t$ - Time, in seconds.

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $t = 15\,s$ , then the angular acceleration of the disc is:

$\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}$

$\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$

The angular acceleration of the disc is $\frac{1}{36}$ radians per square second.

ii) The number of rotations that the disk makes before it stops ( $\Delta \theta$ ), in revolutions, is determined by the following formula:

$\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}$ (2)

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$ , then the number of rotations done by the disc is:

$\Delta \theta = 3.125\,rev$