I need help on this and the first person who will answer this correctly gets BRANLIST

Why is a cell theory a theory and not a law

geniusboy [140]

Laws are very rare in science and they focus on simple, clear, and undisputed facts. A scientific theory is an explanation that is based upon the evidence that is available. With the theory of evolution, and many other scientific theories, the Cell Theory has extremely strong evidence to support it.
Hope this helps!

7 0

2 years ago

Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s

umka2103 [35]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=2.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}$

$\\ \sf\longmapsto Acceleration=1.6m/s^2$

8 0

2 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

Winner of five swimming gold medals at the 1988 olympics

Fed [463]

Matt Biondi..?
(I don’t know if it’s right, sorry if it is wrong)
:)

6 0

2 years ago

HELP ASAP!! WILL MARK BRAINLIEST!! WRITE IN YOUR OWN WORDS!! What is something that you would like to see a physicist develop in

Schach [20]

Answer:

Phones as sunglasses with a mic. I put on my glasses and I say what's the weather today, The sunglasses will tell me the weather and can be charged just like phones

Explanation:

8 0

3 years ago

I need help on this and the first person who will answer this correctly gets BRANLIST​

I need help on this and the first person who will answer this correctly gets BRANLIST