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3 years ago

Match the organisms to the descriptions.

Physics

2 answers:

makvit [3.9K]3 years ago

6 0

Answer:

Lacks colored blood - Scorpion

Soft, unsegmented body - Octopus

A vertebrate - Bird

Lacks antennae - Starfish

Explanation:

Scorpion belongs to the Phylum Arthropoda of the Kingdom Animalia. All the organisms of this phylum lack coloured blood.

Octopus comes under the Phylum Molusca of the Kingdom Animalia. Soft, unsegmented body is the property of the organisms of this phylum.

Bird belongs to the Vertebrata class of the Phylum Chordata of Kingdom Animalia.

Starfish belongs to Echinodermata Phylum of the Kingdom Animalia. The organisms of this category do not posses antennae.

shtirl [24]3 years ago

5 0

Lacks colored blood - Scorpion

Soft, unsegmented body - Octopus

A vertebrate - Bird

Lacks antennae - Starfish

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Read 2 more answers

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Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

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The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$