Buy a fan or heater, get a blanket, take a shower, go outside, go by the fireplace
Answer:
The unit that is a radical among the options is;
a. NH₄⁺
Explanation:
A radical is an atom, group of atoms molecule or ion that contains an electron that is unpaired. The presence of the unpaired electrons make radical very reactive as such most radicals enter into reactions with other species or with themselves to form new compounds making them not last long.
Radicals are also known as free radicals. Examples of radicals includes; Ammonium ion, NH₄⁺
Nitrate ion NO₃⁻ (From a salt)
Hydroxide ion OH⁻ (From a base)
Sulphate ion SO₄²⁺ (From an acid)
All of which are formed in aqueous solution of acids, salts and bases.
Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be:
<span>L = R * m * vi * cos(90 - theta) </span>
<span>cos(90 - theta) is just sin(theta) </span>
<span>and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g </span>
<span>so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g </span>
<span>We can combine the two vi terms and get: </span>
<span>L = vi^3 * m * sin(theta) * sin(2*theta) / g </span>
<span>What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. </span>
<span>Now, for the first part... </span>
<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>
<span>Okay, so let's get back to what we know: </span>
<span>L = d * m * v * cos(phi) </span>
<span>where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... </span>
<span>m is still m. </span>
<span>v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) </span>
<span>d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g </span>
<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>
<span>The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: </span>
<span>h = 1/2 * (vi * sin(theta))^2 / g </span>
<span>So, there's the rise. So, our *slope* is rise/run, so </span>
<span>slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] </span>
<span>The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: </span>
<span>slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] </span>
<span>(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). </span>
<span>Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... </span>
<span>slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 </span>
<span>Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... </span>
<span>alpha = arctan( tan(theta) / 4 ) </span>
<span>Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha </span>
<span>so, phi = 180 - arctan( tan(theta) / 4 ) </span>
<span>Now, we go back to our original formula and plug it ALL in... </span>
<span>L = d * m * v * cos(phi) </span>
<span>becomes... </span>
<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] </span>
<span>Now, cos(180 - something) = cos(something), so we can simplify a little bit... </span>
<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ] </span>
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
<h2>Answer: The separation of the components of the nucleous of the atom </h2><h2>
</h2>
The n<u>uclear fission</u> consists of dividing a heavy nucleus into two or more lighter or smaller nuclei, by means of the <u>bombardment with neutrons to make it unstable.
</u>
Then, with this division a great release of energy occurs and the emission of two or three neutrons, other particles and gamma rays.
It should be noted that in the process, the emitted neutrons can interact with new fissionable nuclei that will emit new neutrons and so on. Effect better known as chain reaction.
