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2 years ago

What are AR,L,K in the periodic table

Physics

2 answers:

Scilla [17]2 years ago

7 0

Answer:

Argon, Lithium and Potassium

Explanation:

Lelu [443]2 years ago

4 0

Answer:

AR=Argon, L=Lithium K=Potassiaum

Explanation:

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At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago

Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di

jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at $20^0\ C$ is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

$d=\dfrac{\lambda}{2}$ ........(1)

Velocity of sound wave is given by :

$v=f\times \lambda$

$d=\dfrac{v}{2f}$

$d=\dfrac{343\ m/s}{2\times 686\ Hz}$

$d=0.25\ m$

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

$d=n\lambda$ ( n = integers )

Let n = 1

So, $d=\dfrac{v}{f}$

$d=\dfrac{343\ m/s}{686\ Hz}$

$d=0.5\ m$

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0

3 years ago

Read 2 more answers

Mike is a runner. Mike runs 200m in 25s; what is his average velocity during the run? assume that units are in m/s. (asap, will

maw [93]

Velocity is displacement/time
(Displacement is the overall change in distance)

So you’ll want to divide 200 by 25, which should give you:

8 m/s

5 0

2 years ago

If five joules were required to move a crate in 3.7 seconds, what power was applied?

AleksAgata [21]

Answer:

The answer to your question is 1.35 Watts

Explanation:

Data

Work = W = 5 J

time = t = 3.7 s

Power = P = ?

Formula

Power is a rate in which work is done or energy is transferred over time

P = $\frac{W}{t}$

Substitution

$P = \frac{5}{3.7}$

Result

P = 1.35 W

8 0

2 years ago

Please help ill mark youas brainliest !!

USPshnik [31]

Answer:

can you put on a clearer image this one is hard to see

8 0

1 year ago