<h2>Amoeba / Unicellular</h2><h2>Segmented worm / Earthworm</h2><h2>Unsegment worm / Tapeworm</h2><h2>Snail / Molluscs</h2><h2>Butterfly / A pair of antenna</h2><h2 /><h3><em>Unicellular: </em><u><em>aboema</em></u><em>: a </em><u><em>one-celled</em></u><em>, microscopic organism belonging to any of several families of rhizopods that move and feed using pseudopodia and reproduce by fission</em></h3><h3><em /></h3><h3><em>Segmented worms: segmented worms include the common </em><u><em>earthworm</em></u><em> and leeches.</em></h3><h3><em /></h3><h3><u><em>Unsegented worms:</em></u><em> unsegmented Worms Phylum Platyhelminthes & Nematoda. Worms. Worms are divided into three different phyla: Phylum Platyhelminthes, the flatworms. These include marine flatworms, flukes, and </em><u><em>tapeworms</em></u><em>.</em></h3><h3><em /></h3><h3><u><em>Molluscs</em></u><em>: molluscs examples: – </em><u><em>snails</em></u><em>, slugs, limpets, whelks, conchs, periwinkles, etc. Class Bivalvia – clams, oysters, mussels, scallops, cockles, shipworms, etc. The Class Scaphopoda contains about 400 species of molluscs called tooth or tusk shells, all of which are marine.</em></h3><h3><em /></h3><h3><u><em>Antennas</em></u><em>: </em><u><em>Nearly all insects have a pair of antennae</em></u><em> on their heads. They use their antennae to touch and smell the world around them. ... Insects are the only arthropods that have wings, and the wings are always attached to the thorax, like the legs.</em></h3>
Answer:
4.19 km and 107.35 degrees north of east
Explanation:
So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:
km
north or west or (180 - 72.65) = 107.35 degrees north of east
Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Answer:
A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.
A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.
An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.
I HOPE THIS IS HELPFUL.
