All categories

3 years ago

which of the following has the greatest impact on an ecosystem migratory animals predation human inference

Physics

1 answer:

laiz [17]3 years ago

6 0

Answer:

human interference

Explanation:

Think of the hybrid lady bug..introduced to control insects on crops.

now they are reproducing rapidly, eating natural ladybugs, and other helpful insects..I haven't seen a red ladybug for years. in nw Illinois..

You might be interested in

What is convection? what is radiation ? what conduction?

stepan [7]

--ⓒᴄᴏɴᴠᴇᴄᴛɪᴏɴ ᴏᴄᴄᴜʀꜱ ᴡʜᴇɴ ᴘᴀʀᴛɪᴄʟᴇꜱ ᴡɪᴛʜ ᴀ ʟᴏᴛ ᴏꜰ ʜᴇᴀᴛ ᴇɴᴇʀɢʏ ɪɴ ᴀ ʟɪQᴜɪᴅ ᴏʀ ɢᴀꜱ ᴍᴏᴠᴇ ᴀɴᴅ ᴛᴀᴋᴇ ᴛʜᴇ ᴘʟᴀᴄᴇ ᴏꜰ ᴘᴀʀᴛɪᴄʟᴇꜱ ᴡɪᴛʜ ʟᴇꜱꜱ ʜᴇᴀᴛ ᴇɴᴇʀɢʏ. ... ʟɪQᴜɪᴅꜱ ᴀɴᴅ ɢᴀꜱᴇꜱ ᴇxᴘᴀɴᴅ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ʜᴇᴀᴛᴇᴅ. ᴛʜɪꜱ ɪꜱ ʙᴇᴄᴀᴜꜱᴇ ᴛʜᴇ ᴘᴀʀᴛɪᴄʟᴇꜱ ɪɴ ʟɪQᴜɪᴅꜱ ᴀɴᴅ ɢᴀꜱᴇꜱ ᴍᴏᴠᴇ ꜰᴀꜱᴛᴇʀ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ʜᴇᴀᴛᴇᴅ ᴛʜᴀɴ ᴛʜᴇʏ ᴅᴏ ᴡʜᴇɴ ᴛʜᴇʏ ᴀʀᴇ ᴄᴏʟᴅ.

--Ⓡʀᴀᴅɪᴀᴛɪᴏɴ ɪꜱ ᴇɴᴇʀɢʏ ᴛʜᴀᴛ ᴄᴏᴍᴇꜱ ꜰʀᴏᴍ ᴀ ꜱᴏᴜʀᴄᴇ ᴀɴᴅ ᴛʀᴀᴠᴇʟꜱ ᴛʜʀᴏᴜɢʜ ꜱᴘᴀᴄᴇ ᴀɴᴅ ᴍᴀʏ ʙᴇ ᴀʙʟᴇ ᴛᴏ ᴘᴇɴᴇᴛʀᴀᴛᴇ ᴠᴀʀɪᴏᴜꜱ ᴍᴀᴛᴇʀɪᴀʟꜱ. ... ᴛʜᴇ ᴋɪɴᴅꜱ ᴏꜰ ʀᴀᴅɪᴀᴛɪᴏɴ ᴀʀᴇ ᴇʟᴇᴄᴛʀᴏᴍᴀɢɴᴇᴛɪᴄ (ʟɪᴋᴇ ʟɪɢʜᴛ) ᴀɴᴅ ᴘᴀʀᴛɪᴄᴜʟᴀᴛᴇ (ɪ.ᴇ., ᴍᴀꜱꜱ ɢɪᴠᴇɴ ᴏꜰꜰ ᴡɪᴛʜ ᴛʜᴇ ᴇɴᴇʀɢʏ ᴏꜰ ᴍᴏᴛɪᴏɴ). ɢᴀᴍᴍᴀ ʀᴀᴅɪᴀᴛɪᴏɴ ᴀɴᴅ x ʀᴀʏꜱ ᴀʀᴇ ᴇxᴀᴍᴘʟᴇꜱ ᴏꜰ ᴇʟᴇᴄᴛʀᴏᴍᴀɢɴᴇᴛɪᴄ ʀᴀᴅɪᴀᴛɪᴏɴ

--Ⓒᴄᴏɴᴅᴜᴄᴛɪᴏɴ ɪꜱ ᴛʜᴇ ᴡᴀʏ ɪɴ ᴡʜɪᴄʜ ᴇɴᴇʀɢʏ ɪꜱ ᴛʀᴀɴꜱꜰᴇʀʀᴇᴅ (ᴛʜʀᴏᴜɢʜ ʜᴇᴀᴛɪɴɢ ʙʏ ᴄᴏɴᴛᴀᴄᴛ) ꜰʀᴏᴍ ᴀ ʜᴏᴛ ʙᴏᴅʏ ᴛᴏ ᴀ ᴄᴏᴏʟᴇʀ ᴏɴᴇ (ᴏʀ ꜰʀᴏᴍ ᴛʜᴇ ʜᴏᴛ ᴘᴀʀᴛ ᴏꜰ ᴀɴ ᴏʙᴊᴇᴄᴛ ᴛᴏ ᴀ ᴄᴏᴏʟᴇʀ ᴘᴀʀᴛ).

3 0

3 years ago

. The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum

Tju [1.3M]

Explanation:

120+60=180

120-60=60

5 0

3 years ago

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+

babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• $\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j$ with $0\le x\le2$ and $0\le y\le6-x$ ;

• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

• $\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k$ with $0\le u\le\frac\pi2$ and $0\le y\le6-2\cos u$ .

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

$\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k$

$\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j$

$\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i$

$\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j$

$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of f with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0$

$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

$\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8$

$\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz$

$=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0$

$\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du$

$=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi$

$\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du$

$=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24$

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV$

where R is the interior of S. We have

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7$

The integral is easily computed in cylindrical coordinates:

$\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2$

$\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3$

as expected.

4 0

3 years ago

Food provide your body with

Levart [38]

The food you eat every day provides the nutrients you need to survive. These food components include the macronutrients – protein, carbohydrate and fat – that offer calories as well as play specific roles in maintaining your health. Micronutrients, such as vitamins and minerals, don’t act as an energy source but do serve a variety of critical functions to ensure your body operates as optimally as possible.

7 0

3 years ago