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

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small

Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0

3 years ago

When flying the LNAV approach, the missed approach point (MAP) would be indicated by reaching __________. A. an altitude of 3100

frutty [35]

Answer:

When flying the LNAV Approach, the missed approach point (MAP) would be indicated by reaching:

C. the RW30 waypoint.

Explanation:

In Aviation, LNAV stands for Lateral Navigation. The option a is incorrect as an altitude of 3100 feet refers to the decision altitude not the missed approach point.
The option b is incorrect as a distance of 1.5 NM to RW30 referring to the Visual descent point (VDP) is 1.5 nautical miles for the Runway (RW) 30 from threshold.
The option c is correct as Missed approach point is designed to coincide with the runway threshold. The RW 30 way point is referring to the way point to the threshold for the Runway 30.

8 0

4 years ago

What is the process called in which water vapor and carbon dioxide retain heat

antoniya [11.8K]

The process in which water vapor and carbon dioxide retain heat is called the greenhouse effect.

<span>There are many gases which causes greenhouse effect like carbon dioxide, methane, nitrous oxide, fluorinated gases etc. These gases trap the heat in the atmosphere, resulting the surface of earth to get warm producing greenhouse effect.</span>

3 0

3 years ago

What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar

Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>

8 0

2 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$