<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:
450
Explanation:
Given,
Mass= 100kg
Velocity= 3 m/s
Kinetic Energy= ?
Kinetic Energy= 1/2 mv^2
= 1/2× 100× 3^2
= 1/2× 900
= 450.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPED</em><em> </em><em>:</em><em>)</em>
Answer:
0.00185 °C
Explanation:
From the question,
The potential energy of the bird = heat gained by the water in the fish tank.
mgh = cm'(Δt)................... Equation 1
Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.
make Δt the subject of the equation
Δt = mgh/cm'............... Equation 2
Given: m = 1 kg, h = 40 m, m' = 50.5 kg
constant: g = 9.8 m/s², c = 4200 J/kg.K
Substitute into equation 2
Δt = 1(40)(9.8)/(50.5×4200)
Δt = 392/212100
Δt = 0.00185 °C
Answer:
T = 295.57 s
Explanation:
given,
mass of the rocket = 200 Kg
mass of the fuel = 100 Kg
acceleration = 35 m/s²
time, t = 35 s
time taken by the rocket to hit the ground, = ?
Final speed of the rocket when fuel is empty
using equation of motion
v = u + a t
v = 0 + 35 x 35
v = 1225 m/s
height of the rocket where fuel is empty
v² = u² + 2 a s
1225² = 0 + 2 x 35 x h₁
h₁ = 21437.5 m
After 35 s the rocket will be moving upward till the final velocity becomes zero.
Now, using equation of motion to find the height after 35 s
v² = u² + 2 g h₂
0² = 1225² + 2 x (-9.8) h₂
h₂ = 76562.5 m
total height = h₁ + h₂
= 76562.5 m + 21437.5 m = 98000 m
now, time taken by before the rocket hit the ground
using equation of motion
negative sign is used because the distance travel by the rocket is downward.
4.9 t² - 1225 t - 13500 = 0
t = 260.57 s
neglecting the negative sign
total time the rocket was in air
T = t₁ + t₂
T = 35 + 260.57
T = 295.57 s
Time for which rocket was in air is equal to 295.57 s.
Answer:
b. $96,914
Explanation:
360-day borrowing rate = 5%
spot rate = 0.48
360-day deposit rate = 6%
Borrow at the rate of 5% to get
SF200,000/1.05 = $190,476.19
Convert at the spot rate of $0.48 to get
190,476.19*0.48 = $91,428.57
Invest at the interest rate of 6% to get
91,428.57/1.06 = 96,914.28
Therefore, Parker Company will receive $96,914 in 360 days.
