Answer:
Since cell division occurs twice during meiosis, one starting cell can produce four gametes (eggs or sperm). In each round of division, cells go through four stages: prophase, metaphase, anaphase, and telophase.
Nuclear energy was not developed. It has existed for as long as time has existed, that is, since the big bang.
The thing that was developed was humans' ability to USE nuclear energy, to do what we want it to do, when we want it to do it.
The reason this was first developed was to bomb the holy beans out of Japan, in order to win World War II.
Today (2020), nine of the world's nations are known to have 14,285 nuclear bombs in storage, for the same general purpose. Seven of these nations are storing 1,170 of these bombs (about 8 percent), and the USA and Russia have all the rest ... 13,035 nuclear bombs.
All nine of these nations promise that they have no plan to use their bombs, they don't want to use them, it would be wrong and terrible to use them, and they will never be the first to use them, but they need to modernize their bombs so that theirs are better than anybody else's bombs, and they need to keep their bombs for as long as anybody else has any, and then maybe a little longer, just in case.
In the years after the ability to bomb the holy beans out of other people was developed, and enough equipment was built to do it 14 thousand times, the ability to use nuclear energy for other purposes was also developed. It's used now to generate electrical energy, and to do several jobs in Medical science.
Answer:
66.85 m
Explanation:
We are given that
Acceleration ,a=
Speed of truck, v=9.5 m/s
We have to find the distance beyond which the traffic signal will the automobile overtake the truck.
Initial speed of automobile, u=0
We know that
Using the formula
For constant speed
Acceleration, a=0
Again
Substitute the value of t
Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m
Answer:
El astrónomo alemán Johannes Kepler
Explanation:
Primera Ley:
Los planetas giran alrededor del Sol siguiendo una trayectoria elíptica. El Sol se sitúa en uno de los focos de la elipse.
La excentricidad e de una elipse es una medida de lo alejado que se encuentran los focos del centro.
Pues bien, la mayoría de las órbitas planetarias tienen un valor muy pequeño de excentricidad, es decir e ≈ 0. Esto significa que, a nivel práctico, pueden considerarse círculos descentrados.
Segunda Ley:
La recta que une el planeta con el Sol barre áreas iguales en tiempos iguales.Para que esto se cumpla, la velocidad del planeta debe aumentar a medida que se acerque al Sol. Esto sugiere la presencia de una fuerza que permite al Sol atraer los planetas, tal y como descubrió Newton años más tarde.
Tercera ley de Kepler:
La tercera ley, también conocida como armónica o de los periodos, relaciona los periodos de los planetas, es decir, lo que tardan en completar una vuelta alrededor del Sol, con sus radios medios.
Para un planeta dado, el cuadrado de su periodo orbital es proporcional al cubo de su distancia media al Sol.
Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.
You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.
The parallel combination of the first two resistors looks like a single
resistor, whose value is
1 / (1/6 + 1/12)
= 1 / (2/12 + 1/12)
= 1 / (3/12)
= 12/3 = 4Ω .
Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of
4Ω + 2Ω = 6Ω .
So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is
I = V / R = (24v) / (6Ω) = 4 Amperes.
That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.
So the current through the 2Ω resistor is 4 Amperes. (B).
Note:
The POWER dissipated by the 2Ω resistor is
P = I² R = (4A)² · (2Ω) = 32 watts .
This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.
