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4 years ago

13

The Clean Air Act of 1963 was an improvement over the Air Pollution Control Act of 1955 because it _____.

1 answer:

Sveta_85 [38]4 years ago

5 0

Was the first act to actually set emissions towards factory smoke and pollution

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a rocket with a lift of mass 20 000kg is blasted upwards with an initial acceleration of 5.0m/s². calculate the initial thurst(f

gavmur [86]

Answer:

Mass of the rocket, m = 20,000 kg Initial acceleration, a = 5 m/s Acceleration due to gravity, g = 10 m/s Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation: F – mg = ma F = m ( g + a ) = 20000 × (10 + 5)

Explanation:

6 0

3 years ago

A thin oil slick (no=1.50) floats on water (nw=1.33). When a beam of white light strikes this film at normal incidence from air,

Sedbober [7]

Answer:

The (minimum) thickness of the oil slick is 325 nm.

Explanation:

Let the (minimum) thickness of the oil slick = $t_{o}$

Therefore:

2 $t_{o}$ = ( $m_{red}$ + $\frac{1}{2}$ )*λ $_{red}$ / $n_{o}$ (1)

similarly,

2 $t_{o}$ =( $m_{violet}+\frac{1}{2}$ )*λ $_{violet}$ / $n_{o}$ (2)

Thus, equation 1 = equation 2

( $m_{red}$ + $\frac{1}{2})$ *λ $_{red} /n_{o}$ = ( $m_{violet}+\frac{1}{2})$ *λ $_{violet}/n_{o}$

Where:

λ $_{red} = 650 nm$

λ $_{violet} = 390 nm$

$n_{o} = 1.5$

Therefore:

$\frac{2m_{violet}+1 }{2n_{red}+1 }$ =650/390=5/3

This shows that,

$m_{violet}$ =2

$m_{red}$ =1

Thus, using equation 2

$t_{o}$ = $\frac{5*390}{2*2*1.5} =325 nm$

6 0

3 years ago

Rate in miles/hours if your run at a speed 2.2 20miles

Vinvika [58]

9.1 miles per hour because 2.2 is your hours right?

5 0

4 years ago

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

4 years ago

Please Help!.....................

Setler79 [48]

The awnser is tranverse wave
hope that helps

4 0

3 years ago