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2 years ago

7

In grassland regions, rainy seasons and drought seasons determine, in part, the _____. kinds of resident organisms spread of fir

es average temperature location of fresh water streams

2 answers:

qaws [65]2 years ago

6 0

kinds of resident organisms

trasher [3.6K]2 years ago

6 0

The correct answer is option A

In grassland regions, rainy seasons the drought seasons determine, in part the kinds of resident organism.

The adaptions of the organism depends on the climatic and environmental conditions in which they live.

By knowing the environmental conditions of the place we get an idea of the organisms that survive there.

You might be interested in

An object has a position given by r = [2.0 m + (2.00 m/s)t] i + [3.0 m − (1.00 m/s^2)t^2] j, where quantities are in SI units. W

lidiya [134]

Answer: 1 m/s

Explanation:

We have an object whose position $r$ is given by a vector, where the components X and Y are identified by the unit vectors $i$ and $j$ (where each unit vector is defined to have a magnitude of exactly one):

$r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j$

On the other hand, velocity is defined as the variation of the position in time:

$V=\frac{dr}{dt}$

This means we have to derive $r$ :

$\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j$

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} t) j$ This is the velocity vector

And when $t=2s$ the velocity vector is:

$\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} (2 s)) j$

$\frac{dr}{dt}=2 m/s i - 1m/s j$ This is the velocity vector at 2 seconds

However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed $S$ :

$S=\sqrt {-1 m/s j + 2 m/s i}$

$S=\sqrt {1 m/s}$

Finally:

$S=1 m/s$ This is the speed of the object at 2 seconds

6 0

1 year ago

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

$34 v + 54 (3 m/s) = 0$

$v = -4.76 m/s$

8 0

1 year ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$

We know that $K_c$ is 33Mpa*m^{0.5} and our Safety factor is 2,

$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

$F_{allow}=\sigma_{allow}*L_c*W$

$F_{allow} = 208.15*250*4$

$F_{allow} = 208.15kN$

That is the maximum tensile load before a catastrophic failure.

4 0

1 year ago

What is the absolute value of the horizontal force that each athlete exerts against the ground?

alexandr402 [8]

Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.

6 0

1 year ago

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?

kaheart [24]

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

F= 0.034 N

Explanation:

Given Data

Outer=2 cm

Edge of blade=21 cm

Mass=7.7 g

Length of blade=21 cm

The last 2cm is treated as if they were at a point 20cm from the center of rotation

To Find

Force=?

Solution

Convert the given frequency to angular frequency

ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)

ω= 3/2*π rad/sec

Now to find centripetal force.

F = m×v²/r

F= m×ω²×r

Put the data

F = 0.0077 kg × (3/2×π rad/sec )²× 0.20 m

F= 0.034 N

3 0

1 year ago