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3 years ago

Which of the following terms helps to describe an ecosystem?

1 answer:

4 0

The answer is all of the above because an ecosystem is how organisms and their environment interact. Population is animals, niche is and organism, and the habitat is the environment.

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Matter is anything that...

Lisa [10]

The answer is b, anything that has mass and takes up space

8 0

3 years ago

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What is the velocity of a wave that has a frequency of 400Hz and a wavelength of 0.5 meters

iren2701 [21]

Answer: 12

Explanation:

Let’s take for instance the case of a wave with a frequency of 400 Hz going through a material at a speed of .5 m/s. The wavelength result is 12 m. Wave velocity (m/s) = Frequency (Hz) x Wavelength (m)

4 0

3 years ago

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A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?

EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by $d = \frac { m } { v }$ .

Here in this problem, we are given a mass of $565 g$ which occupies a volume of $50 cm^3$ .

So plugging the data in the above formula to find the density:

Density = $\frac { 565 } { 50 } = 11.3$

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

8 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .