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3 years ago

How did Linnaeus improve the classification of organisms

Physics

2 answers:

mamaluj [8]3 years ago

6 0

Answer: A

Reason: one of his contributions was the development of a hierarchical system of classification of nature. Linnaeus also provided us with a consistent way to name species called binomial nomenclature. (From Internet)

Darina [25.2K]3 years ago

5 0

Answer: my screen looks exactly the same and i was looking for the answer swell but i noticed that the class matches mines as well,im trying to say i think that we know each other plz reply if you go to Georgia jones ayers middle school, and have teacher name ms.banks

Explanation:

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A person runs 15 kilometers in two hours what is his/her average speed

bazaltina [42]

Answer:

$\displaystyle V=7.5\ km/h$

Explanation:

Average Speed

If an object travels a distance d in a time t regardless of the direction, the average speed is the quotient of the distance over the time:

$\displaystyle V=\frac{d}{t}$

It's known a person runs d=15 kilometers in t=2 hours, thus his/her average speed is:

$\displaystyle V=\frac{15\ km}{2\ h}$

Calculating:

$\boxed{\displaystyle V=7.5\ km/h}$

5 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:

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

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

Read 2 more answers

Density is ———- Through the world’s ocean

Montano1993 [528]

Density is varying through the world’s oceans. The density of ocean water is not uniform.

7 0

4 years ago

Question 1

Alexus [3.1K]

Question 1's answer is, if we change the length of the string allowed to vibrate, then the frequency (and pitch) of the string vibration will change. This is because the tension of the string determines the pitch.

Question 2's answer is 2. This can be solved by doing simple division. 30/15 = 2.

Question 3's answer is short bars. Each bar on a xylophone produces a different pitch; the shorter the bar, the higher the pitch.

Hope this helps.

7 0

3 years ago

Which option is a force: A. velocity B. acceleration C. mass D. weight

ehidna [41]

Answer:

Explanation:

Weight and force have same derived units

8 0

3 years ago