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2 years ago

Reduction of the amount of waste entering the waste stream is called __________

1 answer:

3 0

The answer that fits the blank is SOURCE REDUCTION. Source reduction refers to the prevention of further pollution by eliminating the source before this can produce the wastes. This helps in the reduction of pollution, thus reduces the amount of toxins brought by these wastes that may harm the environment.

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Which of these statements are true?

Juliette [100K]

D.Neither High- nor low-pressure systems lead to rain

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2 years ago

Read 2 more answers

An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate

Bumek [7]

Answer:

a) n = 9.9 b) E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

In this case En = 19 eV let us reduce to the SI system

En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

n = √ (In 8 m L² / h²)

let's calculate

n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

n = √ (98) n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

In = (h² / 8mL2² n²

In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²

E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

E₁₀ = 1.925 101 eV

E₁₀ = 19.25 eV

the result with significant figures is

E₁₀ = 19.25 eV

3 0

2 years ago

8. The legs of a young man are each 0.650 meters long. What is his maximum walking speed?

Norma-Jean [14]

Answer:

2.52 m/s

Explanation:

When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.

Sum of the forces in the radial direction:

∑F = ma

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 0.650 m:

v = √(9.8 m/s² × 0.650 m)

v = 2.52 m/s

8 0

2 years ago

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

4 0

2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

2 years ago