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irinina [24]
3 years ago
8

Which method involves using tools to control invasive species?

Physics
1 answer:
vitfil [10]3 years ago
5 0

Answer;

Mechanical control

Explanation;

-Invasive species may cause environmental harm, economic harm, or impact human health. A key factor that makes many species invasive is a lack of predators in the new environment.

-Mechanical control usually refers to the mowing or mechanical cutting of an invasive plant infestation to limit seed production. With mowing, timing is essential. Invasive plants must be removed before the plants go to seed in order to be an effective method of control. Plants should be cut as close to the ground as possible and may have to be treated more than once in a growing season to achieve desired results.

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Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s
Daniel [21]

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

v_f^2 - v_i^2 = 2 ad

so here we have

0^2 - 40^2 = 2 a (70 feet)

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

0^2 - 48^2 = 2 a (d)

now divide the above two equations

\frac{40^2}{48^2} = \frac{70 feet}{d}

d = 100.8 ft

4 0
3 years ago
What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?
Andru [333]
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26} protons

Then, the total charge is the number of protons times the charge of a  single proton:
Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C
8 0
3 years ago
Read 2 more answers
Why do most amphibians return to the water to reproduce?
Ymorist [56]
C. amphibian eggs do not contain a protective shell
5 0
2 years ago
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A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
3 years ago
‏A 50 - N x m torque acts on a wheel with a moment of inertia 150 kg x m² . If the wheel starts from rest , how long will it tak
denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, \triangle \theta = 2\pi

Torque, \tau = 50 Nm

Moment of Inertia, I = 150 kg m^2

If the wheel starts from rest, w_{0} = 0 rad/s

The angular displacement of the wheel can be given by the formula:

\triangle \theta = \omega_0 t + 0.5 \alpha t^2................(1)

Where \alpha is the angular acceleration

\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2

To get t, put all necessary parameters into equation (1)

2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s

3 0
3 years ago
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