Jellyfish reproduction<span> involves several different stages. In the adult, or medusa, stage of a </span>jellyfish<span>, they can </span>reproduce<span>sexually by releasing sperm and eggs into the water, forming a planula. ... During this stage, which can last for several months or years, asexual </span>reproduction<span> occurs.</span>
Answer:
D metallic
Explanation:
The chemical bonding which rises from electrostatic attractive force between the conduction electrons and the positively charged metal ions is called metallic bonding.
<u>It is sharing of the free electrons among the structure of the positively charged ions which are known as cations.
</u>
<u>In this type of bonding, these free electrons freely move in the crystal mattice of the metal. </u>
The bonding accounts for properties of metals, such as ductility, strength, electrical and thermal conductivity and resistivity and luster.
<span>The solubility of KClO</span>₃ : ( 10.1 / 100 g water ) at 30ºC
10.1 g ------------ 100 g ( H₂O )
? g ------------- 100 g ( H₂O )
Mass of KClO₃ :
100 * 10.1 / 100
1010 / 100 = 10.1 g of KClO₃
hope this helps!
The states of matter is solid,liquid and gas & you go from there.
Answer:
the stronger light 5.5 m apart from the total illumination
Explanation:
From the problem's statement , the following equation can be deducted:
I= k/r²
where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality
denoting 1 as the stronger light and 2 as the weaker light
I₁= k/r₁²
I₂= k/r₂²
dividing both equations
I₂/I₁ = r₁²/r₂²=(r₁/r₂)²
solving for r₁
r₁ = r₂ * √(I₂/I₁)
since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus
r₂ = r₁ + d
then
r₁ = (r₁ + d)* √(I₂/I₁)
r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)
r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)
r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =
r₁ = d/[√(I₁/I₂)-1)]
since the stronger light is 9 times more intense than the weaker
I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3
then since d=11 m
r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m
r₁ = 5.5 m
therefore the stronger light 5.5 m apart from the total illumination
