The answer is 30.
Explanation:
4m•(3)=12m
10•(3)=30
When the number of electrons striking the anode of an x-ray tube is increased, the <u>density</u> of emitted X-Rays increases.
Option: 1
<u>Explanation:</u>
As the electron speed increases, the heat radiation also increases from thermionic emission, which causes more heat and more X-ray release. X-rays are produced by an a vacuum tube called X-ray tube that uses more voltage to make the electrons accelerate which the hot cathode releases to a high velocity.
This high speed electrons meets in a collision with a metal target which is the anode, and thus create the X-rays. So, the electron number available and the time period set for their release from the filament determines how many x-rays are produced from the anode. Hence, more the number of electrons striking the anode,the more is the emission of x-rays.
To find the height of the cliff
use the kinematical equation
height = 1/2*a*t^2
= 5*0.75*0.75 = 2.8125 m
Answer:
The frequency of radio waves with a wavelength of 20m is
1.5 e+7 Hertz (Hz) or 15 million Hertz.
This was obtained from the formula speed (c) is equal to the product of frequency (f) and wavelength (λ)
c = fλ
Since we are looking for the frequency, we will divide both sides of the formula above by the lamda symbol for wavelength, which gives us:
f = c / λ
The value of wavelength is already given which is 20 meters. The speed of radio waves is the same with the speed of light when travelling in a vacuum – also with the other electromagnetic waves. Hence the speed (c) is 3 e+8 m/s
f = 3 e+8 m/s / 20m
f = 1.5 e+7 H
Explanation:
this is the correct answer you want.
(a) we can solve the problem by using the mirror equation:
where
f is the focal length of the mirror
is the distance of the object from the mirror
is the distance of the image from the mirror.
For the sign convention, f is taken as negative for a convex mirror and
is taken as negative if the image is located behind the mirror, as in this problem. So we have
and re-arranging the mirror equation we can find the distance of the object from the mirror:
from which we find
(b) The magnitude is defined as the ratio between the size of the image and the size of the object, which is also equal to the negative of the ratio between the distance of the image and the distance of the object from the mirror:
Using what we found at point (a), the magnification in this problem is
where the positive sign means the image is upright.