Answer:
1985kg
Explanation:
assuming that
pi =3.14
oil density = 950kg/ cubic meter
g= 9.8m/s

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:

Where,
F =View Factor
A = Cross sectional Area
Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m

The view factor between two coaxial parallel disks would be


Then the view factor between base to top surface of the cylinder becomes
. From the summation rule


Then the net rate of radiation heat transfer from the disks to the environment is calculated as





Therefore the rate heat radiation is 780.76W
Answer:
Going by EM SPECTRUM WE HAVE
radio waves, microwaves, infrared, VISIBLE LIGHT, ultraviolet, X-rays, GAMMA RAYS
Explanation:
BECAUSE
V= WAVELENGTH/ FREQUENCY
AS FREQUENCY INCREASES WAVELENGTH DECREASE AN VICE VERSA
Alkali metals: left column of your periodic table (not hydrogen, but anything below it). They have one valence electron, which they are happy to share in a reaction.
Halogens: second column from the right of your periodic table. They are one electron short of a full shell, so they are reactive in the opposite way that alkalis are--they want electrons.
Atomic number (number of protons) is the big number on the periodic table square. Hydrogen's is 1.
Atomic mass is a little number down below. For example, Hydrogen's is 1.008.
Neutrons are a tricky subject, because different isotopes of the same element can have different numbers of neutrons. You can't generally get this from the atomic mass, because the atomic mass is a weighted average of naturally occurring isotopes. Hydrogen can have 0,1, or 2 neutrons. To answer this, you'd have to choose a particular isotope from the table of isotopes (a completely different chart from the periodic table) which has a certain number of neutrons: n = weight - Z.
Valence electrons are the electrons in the outermost shell. (The column of the table).
<span>
Number of principal shells is the row of the periodic table. </span>
When Object is at zero height, and there is no potential energy possess by the object then it exerts Greatest Kinetic energy in it's whole Journey
Hope this helps!