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What would happen if rats was eliminated from the environment

I am Lyosha [343]

Then the workers that made poison for rats would have to throw all of the poison in the garbage, sad, I know. Oh, if rats was eliminated from the environment, then cats that LOVE rats wont be poisoned by the rats that EAT the poison :). Yes, one of my mama cats died );

4 0

3 years ago

What is the purpose of control group ?

kirill [66]

Control group in a scientific experiment is a group separated from the rest of the experiment, it is where the independent variable being tested cannot influence the results. It isolates the independent variable's effects on the experiment and can helps rule out alternative explanations of the experimental results.

6 0

3 years ago

Suppose you wanted to use a non-reflecting layer for radar waves to make an aircraft invisible. What would the thickness of the

Xelga [282]

Answer:

the thickness of the film for destructive interference is 1 cm

Explanation:

We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when

* the wave passes from the air to the film with a higher refractive index

* the wavelength inside the film changes by the refractive index

λ = λ₀ / n

so the ratio for destructive interference is

2 n t = m λ

t = m λ / 2n

indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness

t = 1 2/2 n

t = 1 / n

where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1

t = 1 cm

So the thickness of the film for destructive interference is 1 cm

8 0

3 years ago

In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf, having about the same mass as it doe

masya89 [10]

The formula of density is given by

Density = Mass ÷ Volume

We have:
Mass = 1.989 × 10³⁰ kg
Volume = $\frac{4}{3} \pi r^3$ = $\frac{4}{3} \pi (7500)^3 = 1.767*10^12$ km³

Density = $\frac{(1.989)(10^{30})}{(1.767)(10^{12})}$ =1.13×10¹⁸ kg/km³

Converting 1.13 × 10¹⁸ kg/km³ to g/cm³

1.13 × 10¹⁸ kg = 1.13 × 10¹⁸ × 10³ = 1.13 × 10²¹ grams
1 km³ = 1 × 10⁶ cm³

(1.13 × 10²¹) ÷ 10⁶ = 1.13 × 10¹⁵ gr/cm³

Answer: Density 1.13 × 10¹⁵ gr/cm³

8 0

4 years ago

Read 2 more answers

The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t

KatRina [158]

Answer:

T₂ = 123.9 N, θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

T₁ₓ - $T_{2x}$ = 0

T_{2x}= T₁ₓ

Y axis y

T_{1y} + T_{2y} - 200N = 0

T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

sin 60 = T_{1y} / T₁

cos 60 = t₁ₓ / T₁

T_{1y} = T₁ sin 60

T1x = T₁ cos 60

T_{1y}y = 100 sin 60 = 86.6 N

T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

T_{2y} = T₂ sin θ

T₂ₓ = T₂ cos θ

we substitute

T₂ sin θ= 200 - 86.6 = 113.4

T₂ cos θ = 50 (1)

to solve the system we divide the two equations

tan θ = 113.4 / 50

θ = tan⁻¹ 2,268

θ = 66.2º

we caption in equation 1

T₂ cos 66.2 = 50

T₂ = 50 / cos 66.2

T₂ = 123.9 N

8 0

3 years ago