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2 years ago

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A group of animals have short, narrow legs that are partly webbed. In which environment do these animals MOST LIKELY live? A) de

serts B) grasslands C) mountains D) rainforests

2 answers:

Burka [1]2 years ago

7 0

I'd say it's answer D) rainforests

Galina-37 [17]2 years ago

4 0

Answer:

D) rain-forests

Explanation:

Such animals live in rain-forests as they need to travel on land as well as in shallow waters. The geography of rain forests is such that it receives a lot of rain throughout the year. This makes it necessary for the animals to have such features in their legs to help them survive in that environment.

For other environment they would not need partly webbed legs.

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Sergeu [11.5K]

The vertical component of the initial velocity is $v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is $v_0_x = \frac{x}{t}$

The horizontal displacement when the object reaches maximum height is $X = \frac{xy}{gt^2} + \frac{x}{2}$

The given parameters;

the horizontal displacement of the object, = x

the vertical displacement of the object, = y

acceleration due to gravity, = g

time of motion, = t

The vertical component of the initial velocity is given as;

$y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt$

The horizontal component of the initial velocity is calculated as;

$x = v_0_xt\\\\v_0_x = \frac{x}{t}$

The time to reach to the maximum height is calculated as;

$T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t$

The horizontal displacement when the object reaches maximum height is calculated as;

$X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}$

Learn more here: brainly.com/question/20689870

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2 years ago

Kristina demonstrates a toy for her younger brother. The steps to operate the toy are listed below. Step 1. Push the toy down. S

IRINA_888 [86]

The object has potential and kinetic energy when it jumps into the air

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3 years ago

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This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

$P = \frac{1}{2} \rho v^3 A C_p$

where $\rho = 1.2 kg/m^3$ is the air density, v = 8.89 m/s is the wind speed, A is the swept area and $C_p = 0.45$ is the efficiency

$10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45$

$10000 = 190A$

$A = 10000 / 190 = 52.7 m^2$

The swept area is a circle with radius r being the blade length

$\pi r^2 = A = 52.7$

$r^2 = 52.7 / \pi = 16.79$

$r = \sqrt{16.79} = 4.1 m$

4 0

2 years ago

Helppppp pleaseee :(

Elza [17]

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

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2 years ago

A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E

Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, $F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}$

Where $G=6.67*10^{-11}$ and $M=5.94*10^{24}$

$F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}$

F= 995.01142 then rounded off

F=995N

6 0

2 years ago