Initia Velocity(u) :- 15m/s
Acceleration(a) :- 4m/s2
Time Taken(t) :- 50 Seconds
{ok a formula is there —->>> s = ut + 1/2 at2(it is ‘t’ square)….. now we will write like this and yeah don’t write this}
We Know,
s = ut + 1/2 at2
s = (15*50 + 1/2 *4*50*50)m
s= (750 + 5000)m
s = 5750 m
Therefore distance traveled is 5750 meters.
Hopefully this helps
They all have instruments to "uncouple" oxidative phosphorylation from electron transport framework by giving an option system to protons to come back to the mitochondrial grid. As protons enter the lattice without going through ATP synthase, their vitality is discharged as warmth. So these produce warm by uncoupling those two procedures.
Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
The lower invertebrate phylum that has the greatest diversity is called Arthropod. An Arthropod is an invertebrate animal that has a segmented body, an exoskeleton, and paired jointed appendages. Arthropods include an incredibly diverse group from the phylum Euarthropoda, which includes myriapods, <span>arachnids, insects, and crustaceans.</span>
I think it 340 living in the wild and145 living in captivity
