Ask question

Ask question

All categories

3 years ago

11

Which of the following is true?

1 answer:

fenix001 [56]3 years ago

8 0

Answer:

Its either A. Or C cause ive had a question like this before So Im sure But if not Then Im so sorry

You might be interested in

Which element contains a full 2p orbital in its valence shell

alex41 [277]

So Neon ( Ne) is the correct answer.

3 0

3 years ago

Read 2 more answers

Which statement is true? A Displacement can never be greater than distance. B Displacement can never be less than distance. C Di

qwelly [4]

Answer:

__________________

I think the answer is A.

__________________

3 0

3 years ago

the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi

leonid [27]

Answer:24.47

Explanation:

Given

$L_1=30 m$

$d_1=3 mm$

$R_1=25 \Omega$

$L_2=40 m$

$d_2=3.5 mm$

we know Resistance $R=\frac{\rho L}{A}$

Where R=resistance

$\rho =resistivity$

L=Length

A=area of cross-section

$A=\frac{\pi d^2}{4}$

Thus $R\propto \frac{L}{d^2}$

therefore

$R_1\propto \frac{L_1}{d_1^2}$ --------1

$R_2\propto \frac{L_2}{d_2^2}$ --------2

divide 1 and 2 we get

$\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}$

$\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}$

$R_2=25\times 0.734\times 1.33$

$R_2=24.47 \Omega$

8 0

3 years ago

A particle’s position is ~r = (ct2 − 2dt3 )iˆ+ (2ct2 − dt2 )jˆ, where c and d are positive constants. Find the expressions for t

Mars2501 [29]

The velocity of the particle is given by the derivative of the position vector:

$\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath$

(a) The particle is moving in the <em>x</em>-direction when the <em>y</em>-component of velocity is zero:

$4ct-2dt = 2t (2c - d) = 0 \implies t=0$

But we want <em>t</em> > 0, so this never happens, unless 2<em>c</em> = <em>d</em> is given, in which case the <em>y</em>-component is always zero.

(b) Similarly, the particle moves in the <em>y</em>-direction when the <em>x</em>-component vanishes:

$2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0$

We drop the zero solution, and we're left with

$c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}$

In the case of 2<em>c</em> = d, this times reduces to <em>t</em> = <em>c</em>/(6<em>c</em>) = 1/6.

7 0

3 years ago

Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up the

Alenkasestr [34]

Answer:

Both warming up and cooling down or not important

8 0

3 years ago