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3 years ago

What is a boyfriend to you? And what age is appropriate to have one?

2 answers:

7 0

A boyfriend is someone who you find more than a friend. Someone who you have a deeper connection with. An age to have one really depends on the person and is mostly decided by parents. If the person is mature they may have one around 15 to 16+ years old. If the person is still "immature" or is not ready to have one, the age limit may be extended to 17+

Andreas93 [3]3 years ago

6 0

Well it depends what your parents say.well you want to find and hang out with someone who is funny a gentlemen. And he respects you.probably helps you put you can follow your heart

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Why should you do you should

uranmaximum [27]

One does not simply should you do you should

8 0

2 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

Which statement explains the polarity of a water molecule?

timofeeve [1]

Answer:

D. The oxygen side is partially negative because electrons are pulled toward the oxygen side.

Explanation:

The water molecule is polar by the virtue of covalent bonds and the hydrogen bonds within and between its molecule.

The oxygen side is partially negative because the electrons are pulled toward the oxygen side.

Between oxygen and hydrogen that makes up the water molecule, oxygen is more electronegative.

An electronegative atom has more affinity for electrons. Since the electrons in the molecule of water is shared between hydrogen and oxygen, the more electronegative specie which is water draws the electron more to itself.

This leaves a net negative charge on the oxygen atom.

8 0

2 years ago

Anyone know the answer ?

nata0808 [166]

Answer:

Explanation:

4 0

2 years ago

Read 2 more answers

In our solar system, most asteroids are found _____.A.In the kuiper beltb.Orbiting close to the sunc.Between mars and jupiterd.B

nydimaria [60]