Answer:
<em>The velocity after 12s is 50.4m/s</em>.
Explanation:
<em>In acceleration formula make velocity the </em><em>subject.</em>
<em> acceleration(a) = velocity(</em>v)÷time(t)
<h3><em> </em><em>velocity</em><em> </em><em>(</em><em>v)</em><em> </em><em>=</em><em> </em><em>acceleration</em><em>(</em><em>a)</em><em>×</em><em>t</em><em>ime</em><em>(</em><em>t)</em></h3>
<em>V </em><em>=</em><em> </em><em>4</em><em>.</em><em>2</em><em>m</em><em>/</em><em>s²</em><em>×</em><em>1</em><em>2</em><em>s</em>
<em>V </em><em>=</em><em> </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s</em>
<em>Therefore</em><em> the</em><em> </em><em>velocity</em><em> </em><em>after</em><em> </em><em>1</em><em>2</em><em>s</em><em> </em><em>is </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s.</em>
I know it’s the Coulomb’s law and that I’m pretty sure the answer would be C.Inverse Square.
Answer:
Ok, primero pensemos en una situación normal.
La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.
Ahora, en este caso tenemos que ignorar los efectos del aire, entonces no hay ninguna fuerza que se oponga a la fuerza gravitatoria, entonces la fuerza neta no cambia a medida que cae (La fuerza neta cambia cuando la moneda impacta el suelo).
También se puede analizar el caso en el que, como la fuerza gravitatoria decrece con el radio al cuadrado, a medida que la moneda cae, la fuerza gravitatoria incrementa. El tema es que en para estas dimensiones, ese cambio en la fuerza gravitacional es imperceptible,
Answer:
becomes narrower
Explanation:
Confidence intervals for the population mean μ and population proportion p becomes narrower as the size of the sample increases.
As,the sample size increases,standard error decreases,so margin of error decreases and hence width of CI decreases.
(I'm lucky to have a computer ... It was only through the miracle of
modern digital technology that I was able to flip your photo right-
side-up to where I could read it.)
Here's how to figure out things like this:
The circle on the left side labeled ' <em>G</em> ' is the <em><u>G</u></em>enerator or battery
that powers this whole circuit and all the devices in it. In order for
any device to work, you need to be able to set your pencil down at
the top of the Generator, and find a path through the circuit and
through that device, where current can flow all the way around to
the bottom of the Generator. If you ever come to an open switch,
then current stops there, and you have to find another way through.
If the path you found takes you back to the bottom of the generator but
it doesn't go through one of the devices, then that device doesn't work.
Look at the picture. If you open switch S-4, then Device-4 can't work,
because current can't go through it from one end of the Generator to
the other end. But all of the other devices still work.
I can see 2 ways to turn off Device-3 with a single switch ... either
open switch S-5, or else open switch S-1. Unfortunately, I think
either way will shut off all 5 devices.