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2 years ago

1. What are the things needed by plants to make their own food?

Physics

1 answer:

iren2701 [21]2 years ago

4 0

Answer:

1-D(carbon dioxide, water and sunlight)

2-D(parasitism)

3-C(competition)

Explanation:

hope it helps

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. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat

AfilCa [17]

Answer:

a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

sin 30 = Wx / W

cos 30 = Wy / W

Wx = W sin30

Wy = W cos 30

Let's write the equations on each axis

X axis

Wx = ma

Y Axis

N- Wy = 0

N = Wy = mg cos 30

N = 2.0 9.8 cos 30

N = 16.97 N

We calculate the acceleration

a = Wx / m

a = mg sin 30 / m

a = g sin 30

a =9.8 sin 30

a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

F -Wx = 0

F = Wx

F = m g sin 30

F = 2.0 9.8 sin 30

F = 9.8 N

5 0

2 years ago

How do very small objects behave?

nataly862011 [7]

A. very small objects behave like like particles.

7 0

3 years ago

Read 2 more answers

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SVETLANKA909090 [29]

Answer:

the rates of rock formation are similar. i could be wrong tho.....

Explanation:

6 0

2 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$