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3 years ago

In which stage in the life cycle of a cnidarian is the organism anchored to the bitten of the ocean or lake where it lives

1 answer:

7 0

<span>In which stage in the life cycle of a cnidarian is the organism anchored to the bitten of the ocean or lake where it lives</span>

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Scientific theories have strod the test of time and will not change

Nataly [62]

That is not a question but not all scientific theories have stood the test of time

8 0

3 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

5 0

3 years ago

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

Zielflug [23.3K]

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

6 0

3 years ago

PLZ HELP

Nadya [2.5K]

1) The mass of the continent is $3.3\cdot 10^{21}kg$

2) The kinetic energy of the continent is 1683 J

3) The speed of the jogger must be 6.57 m/s

Explanation:

The continent can be represented as a slab of size

$d=5850 km = 5.85\cdot 10^6 m$

and depth

$t = 35 km = 3.5\cdot 10^4 m$

So its volume is

$V=d^2 t = (5.85\cdot 10^6)^2(3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that the density of the continent is

$\rho = 2750 kg/m^3$

Therefore, we can calculate its mass as:

$m=\rho V=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

The kinetic energy of the continent is given by

$K=\frac{1}{2}mv^2$

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

The mass is

$m=3.3\cdot 10^{21} kg$

So, the kinetic energy of the continent is

$K=\frac{1}{2}(3.3\cdot 10^{21})(1.01\cdot 10^{-9})^2=1683 J$

Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

$K=\frac{1}{2}mv^2$

where

m = 78 kg is the mass of the jogger

v is his speed

We can therefore re-arrange the equation to find the speed of the man, and we get:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

Does gravity hold things together?

mariarad [96]

<span>Gravity is an attractive force. All masses exert gravity, so matter pull toward each other. However, it is only significant when the masses are large, like that of Earth. Exactly what causes gravity still remains a mystery. Since all the other three fundamental forces (strong, weak, and electromagnetic) are caused by particles called bosons, some scientists think that gravity is caused by particles called the graviton. However, such a particle has not yet been observed, and its existence has not been proven yet.
Also, gravity does not hold EVERYTHING together, as another person has already mentioned. Static electricity holds your socks together after they come out of the dryer, not gravity. Also, gravity is too weak to hold together, say, the nucleus of atoms, or even atoms of molecules together.</span>

3 0

3 years ago

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