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2 years ago

What does biological pest control involve?

1 answer:

5 0

Biological control is a bioeffector-method of controlling pests such as insects, mites, weeds and plant diseases, using other living organisms.
Biological pest control relies on natural mechanisms including predation, parasitism, or herbivory, but typically also involves an active human management role.

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2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the

ASHA 777 [7]

The statement shows a case of rotational motion, in which the disc decelerates at constant rate.

i) The angular acceleration of the disc ( $\alpha$ ), in revolutions per square second, is found by the following kinematic formula:

$\alpha = \frac{\omega_{f}-\omega_{o}}{t}$ (1)

Where:

$\omega_{o}$ - Initial angular speed, in revolutions per second.
$\omega_{f}$ - Final angular speed, in revolutions per second.
$t$ - Time, in seconds.

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $t = 15\,s$ , then the angular acceleration of the disc is:

$\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}$

$\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$

The angular acceleration of the disc is $\frac{1}{36}$ radians per square second.

ii) The number of rotations that the disk makes before it stops ( $\Delta \theta$ ), in revolutions, is determined by the following formula:

$\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}$ (2)

If we know that $\omega_{o} = \frac{5}{12}\,\frac{rev}{s}$ , $\omega_{f} = 0\,\frac{rev}{s}$ y $\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}$ , then the number of rotations done by the disc is:

$\Delta \theta = 3.125\,rev$

The disc makes 3.125 revolutions before it stops.

We kindly invite to check this question on rotational motion: brainly.com/question/23933120

6 0

2 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0

2 years ago

How to do problems 7-10

Stella [2.4K]

I cant see it very clear

8 0

2 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$