Answer:
Δ KE = -495 J
Explanation:
given,
mass of the ice hockey player = 110 Kg
initial speed = 3 m/s
final speed = 0 m/s
distance, d = 0.3 m
change in kinetic energy
Δ KE = -495 J
Hence, the change in kinetic energy is equal to Δ KE = -495 J
<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:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I = ![(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%28mL%5E%7B2%7D%20%29%20%2B%20m%28%5Btex%5D%5Cfrac%7BL%7D%7B2%7D)
)^{2}[/tex]
I = ![(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%280.42%20x%200.75%5E%7B2%7D%20%29%20%2B%20%28%200.42%20x%20%28%5Btex%5D%5Cfrac%7B0.75%7D%7B2%7D)
)^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I
= 0.5 x 0.07875 x 
= 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
solving for t
t = 1.075 sec
for horizontal motion
x = 2cos18*1.075
x = 2.044 m
The correct answer is A. Most reactive non metals.
Firstly if some one knows how to read the periodic table he would have no confusion in deciding whether group 7 has metals or non metals. Group 7 contains non metals so basically we can easily cancel out two options of metals.
Secondly group 7 non metals are the most reactive non metals as they need only one electron in order to complete their valence shell and become stable.
