How can a machine make work easier for you?

Describe and give an example of mutualism.

Slav-nsk [51]

Answer:

Mutualism, commensalism, parasitism, competition, and predation.

Explanation:

mutualism- relationship between two or more organisms where both are benefited. Example-oxpecker with rhino/zebra. They eat bugs off of them which means that they are getting food, while the rhino/zebra are getting cleaned up with pest control.

commensalism- relationship between two organisms where one benefits and the other isn't benefited or harmed. EX- tree frogs use plants as protectioin.he frog is benefited, and the plant is neither harmed nor benefited. Remora fish have a disk on their heads that they use to attach themselves to larger animals for protection. The animals they attach to are neither harmed nor benefited.

parasitism- in a relationship where an organism benefits at the expense of the other. (one is benefited while the other is harmed) ex- fleas and ticks that live on cats and dogs, tape worms that live in people and animals that eat the food which means that the people aren't getting the food or nutrition that they eat. lice, etc

competition- interaction within organisms/species in which both the organisms/species are harmed and is apart of natural selection. Examples may include two males fighting over a mate, animals competing over food, limited habitats that they are fighting over, territory, etc.

predation- the preying of one animal on another. It's where the predator hunts and eats another organism which is its prey. categorized within-(1) carnivory, (2) herbivory, (3) parasitism, and (4) mutualism. Each type of predation can by categorized based on whether or not it results in the death of the prey.ex- owls hunting mice, wolves hunting rabbits, lion hunting gazelle, etc.

7 0

3 years ago

A resistor R and another resistor 2R are connected in a series across a battery. If heat is produced at a rate of 10W in R, then

alekssr [168]

Answer:

Option B

Solution:

As per the question:

Heat produced at the rate of 10 W

The resistor R and 2R are in series.

Also, in series, same current, I' passes through each element in the circuit.

Therefore, current is constant in series.

Also,

Power, $P' = I'^{2}R$

When current, I' is constant, then

P' ∝ R

Thus

$\frac{P'}{2R} = \frac{10}{R}$

P' = 20 W

5 0

3 years ago

The relationship among mass force and acceleration is explained by

Anettt [7]

Newton's second law of motion. F = m a .

7 0

3 years ago

Read 2 more answers

With what minimum speed must you toss a 200 g ball straight up to just touch the 12-m-high roof of the gymnasium if you release

masha68 [24]

Answer:

v = 14.41 m/s

Explanation:

It is given that,

mass of the ball, m = 200 g = 0.2 kg

Height of the roof, h = 12 m

The ball is tossed 1.4 m above the ground, h' = 1.4 m

Let v is the minimum speed with which the ball is tossed. Using the conservation of energy to find it as :

$E_p=E_k$

$mg(h-h')=\dfrac{1}{2}mv^2$

$g(h-h')=\dfrac{1}{2}v^2$

$v=\sqrt{2g(h-h')}$

$v=\sqrt{2\times 9.8(12-1.4)}$

v = 14.41 m/s

So, the minimum speed with which the ball is thrown straight up is 14.41 m/s. Hence, this is the required solution.

4 0

3 years ago

A golf ball is given an initial velocity of 100 m/s at an angle of 22o with the horizontal. The golf ball lands on the roof of a

hoa [83]

a) 92.7 m/s, 37.5 m/s

The initial horizontal and vertical velocity are given by the equations:

$u_x = u cos \theta\\u_y = u sin \theta$

where

u = 100 m/s is the initial velocity of the ball

$\theta=22^{\circ}$ is the angle of projection

So, the horizontal component of the initial velocity is

$u_x = (100)(cos 22)=92.7 m/s$

While the vertical component is

$u_y = (100)(sin 22)=37.5 m/s$

b) 3.83 s

The time at which the ball reaches the maximum height can be found by analyzing the vertical motion only.

In fact, the vertical velocity of the ball is given by

$v_y = u_y +at$

where

$u_y = 37.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (in the downward direction)

t is the time

The ball reaches the maximum height when the vertical velocity becomes zero:

$v_y = 0$

So, solving for t, we find the time at which this happens:

$t=-\frac{u_y}{g}=-\frac{37.5}{9.8}=3.83 s$

c) 92.7 m/s

As stated in part b), the vertical velocity of the ball at the maximum height is zero:

$v_y = 0$

Along the horizontal direction, insteand, the motion of the ball is uniform, since there are no forces acting in this direction. Therefore, the horizontal velocity of the ball is constant (because the horizontal acceleration is zero), and it is

$v_x = u_x = 92.7 m/s$

And therefore, the velocity of the ball at the maximum height is simply equal to the horizontal velocity:

$v=92.7 m/s$

d) 71.7 m

The maximum height of the ball can be found by studying again the vertical motion only, by using the following suvat equation:

$s=u_yt+\frac{1}{2}at^2$

where

s is the vertical displacement at time t

$u_y = 37.5 m/s$

$a=g=-9.8 m/s^2$

We know that the ball reaches the maximum height when t = 3.83 s, so we can find the vertical displacement at that time:

$s=(37.5)(3.83)+\frac{1}{2}(-9.8)(3.83)^2=71.7 m$

So, the maximum heigth is 71.7 m.

e) 5.93 s

We know that the ball lands on a building whose height is

h = 50 m

This means that when the ball lands, its vertical displacement is

s = 50 m

So, we can use again the equation

$s=u_yt+\frac{1}{2}at^2$

To find t, the time at which the ball lands. Substituting:

$u_y = 37.5 m/s$

$a=g=-9.8 m/s^2$

s = 50 m

The equation becomes

$50=37.5 t -4.9 t^2\\4.9t^2 -37.5 t +50 = 0$

which gives two solutions:

t = 1.72 s

t = 5.93 s

The first solution is the moment at which the ball reaches the height of h = 50 m for the first time (before reaching its maximum height), so we ignore it, and therefore the ball lands after

t = 5.93 s

8 0

3 years ago