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Yakvenalex [24]

2 years ago

13

The blue gill is a type of fish that lives in Indiana ponds. It's diet consists of insects snails other small fish and fish eggs

. What role does the bluegill play in the food web of a pond? (And i couldn't find science so i put it under physics)

1 answer:

laiz [17]2 years ago

6 0

Answer:

Bluegills are the primary food source in every farm pond. They spawn from May to September and provide a large amount of food to the predators in the pond.

Explanation:

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A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

5 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

How do you solve <img src="https://tex.z-dn.net/?f=4x%5E%7B3%7D" id="TexFormula1" title="4x^{3}" alt="4x^{3}" align="absmiddle"

Vlad1618 [11]

$Hello$ $There!$

Here's a explanation!

Let's solve your equation step-by-step.

$4x^3=2x^-^1$

$4x^3=\frac{2}{x}$

Step 1: Multiply both sides by x.

$4x^4=2$

$\frac{4x^4}{4} =\frac{2}{4}$

(Divide both sides by 4).

$x^4=\frac{1}{2}$

$x=+(\frac{1}{2} )^(^\frac{1}{4} ^)$

Take the root.

ANSWER!

$x=0.840896$ $Or$ $x=-0.840896$

Hopefully, this helps you!!

$AnimeVines$

8 0

3 years ago

A. Draw four rays parallel to the optical axis of your mirror. Two above the optic axis and two below it.

Katen [24]

Answer:

f = q

Explanation:

In the attachment we can see a diagram of the parallel rays.

The dotted line represents the normal to the mirror surface

These rays when reflected using the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and the image respectively.

Since the rays are parallel P = inf

1 / f = 1 / inf + 1 / q

f = q

this means that all the rays focus on one focal point.

6 0

3 years ago

What is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

Arlecino [84]

The go-kart's velocity $v$ after time $t$ is given by

$v=v_0+at$

where $v_0$ is its initial velocity and $a$ is its acceleration. After $t=5\,\mathrm s$ , the go-kart stops completely, so

$0\,\dfrac{\mathrm m}{\mathrm s}=v_0+\left(-0.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5\,\mathrm s)\implies v_0=2.5\,\dfrac{\mathrm m}{\mathrm s}$

where $a$ because we know the go-kart is slowing down.

7 0

3 years ago