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3 years ago

What happened to the kelp forest when the otter was hunted to near extinction?

1 answer:

7 0

Answer: Sea otter is the pioneer species in the kelp forest as it regulates and controls the population of other species in the kelp forest.

Explanation:

If sea otters are hunted and their population is brought to extinction then this will cause major harm the ecosystem of the kelp forest and it will disturb the ecological balance in the kelp forest. The herbivorous animals consumed by the sea otters will increase in population and they will consume a lot of vegetation in the forest. The kelp forest which forms the coastline will not remain effective in providing protection against the storms to the neighboring areas.

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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

7 0

3 years ago

What is the motional kinetic energy of a 25 kg object moving at a speed of 10 m/s?

Soloha48 [4]

1250kgm²/s is the motional kinetic energy of a 25kg object moving at a speed of 10m/s

Kinetic energy of an object is defined as the energy which is possessed when that is in motion. It is the energy of the kinetic mass of an object. Kinetic energy is never negative and is a scalar quantity. That is, it shows only size, not orientation.

Given to us

Mass of the object, m=25kg

Velocity of the object, v=10m/s

K.E=1/2x25x10²

=1250

Kinetic energy is directly proportional to the mass and velocity squared (K.E.) of an object. =1/2xMxV². If the mass is in kilograms and the velocity is in meters/second, then the kinetic energy is in kilograms - meters squared/second.

Learn ore about Kinetic energy here brainly.com/question/25959744

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5 0

9 months ago

Which ion channels mediate the falling phase of an action potontial?

White raven [17]

Answer:

Voltage-gated K+ channels

4 0

3 years ago

A truck traveled 400 meters north in 80 seconds and then it traveled 300 meters east in 70 seconds. The magnitude of the average

Varvara68 [4.7K]

Answer:

3.3m/s

Explanation:

You first get the total time (80 + 70 = 150s).

Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.

d= 500m.

So now that you have displacement and time, you can find the velocity:

v=d/t

v=500/150

v=3.3

5 0

3 years ago

A 62kg box is lifted 12 meters off the ground. How much work is done?

Temka [501]

Answer: 7291.2 joules

Explanation:

Work is done when force is applied on an object over a distance.

Thus, Workdone = Force X distance

Since Distance moved by box = 12 metres

mass of box = 62kg

Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2

Recall that Force = Mass x acceleration due to gravity

i.e Force = 62kg x 9.8m/s^2

= 607.6 Newton

So, Workdone = Force X Distance

Workdone = 607.6 Newton X 12 metres

Workdone = 7291.2 joules

Thus, 7291.2 joules of work was done.

4 0

3 years ago