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2 years ago

What is the term that we use to describe two species living together?

1 answer:

5 0

D. Symbiosis. Camouflage is a creature disguising itself as its surroundings. Hibernation is a animal sleeping for an extended period of time when food is scarce to conserve energy. Migration is an animal moving from one place to another when food is scarce in the first area so it can find more food in the new area. Symbiosis is two animals living together and helping each other live.

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A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$

Finally, the total work done on the moving particle can be calculated.

$W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J$

4 0

2 years ago

Read 2 more answers

Among the elements potassium, lithium, and iron, the metallic bonds are likely to be strongest in

Mrac [35]

彼が好きな男のような話は妻と関係があったのでそう

Explanation:これが難しい場合はコメントで教えてくださいが何でも

lol okay

from: a random person :)

4 0

2 years ago

HELP! FOR A TEST!

Sati [7]

Answer:

<h3>On a velocity vs time graph the slope of the line represents the acceleration of the object. With a slope of zero, the object is moving at a constant velocity in the positive (+) direction during this five minute interval. ... Displacement and distance can both be determine on a velocity vs.</h3>

4 0

2 years ago

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A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

sleet_krkn [62]

Answer:

v = $\left[\begin{array}{c}0.66&0\end{array}\right]$ m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

$\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}$

The velocity vector v is the first derivative of the position vector r with respect to time:

$\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}$

The given values are:

$t=\frac{x}{v}=\frac{14}{3.8}=3.7 s$

$\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}$

8 0

2 years ago

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, $\omega _i$ = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0

2 years ago