Ask question

Ask question

All categories

Svetradugi [14.3K]

2 years ago

11

What do neon,oxygen and nitrogen have in common?

1 answer:

julsineya [31]2 years ago

7 0

Their Period number is common means their "Principal Quantum Numbers" are same

Hope this helps!

You might be interested in

Formula of relative displacement

Zolol [24]

what is the question?

8 0

3 years ago

Why is venus called brightest star and mars called red planet

bogdanovich [222]

Answer:given below by him is correct

Explanation:

Pls refer his

3 0

2 years ago

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

2 years ago

2. What do pitch and loudness have in common?

DanielleElmas [232]

Answer:

Both are subject to a persons interpretation

Explanation:

We hear people describe this when somebody is making an irresistible sound. usually people say the baby has a pitch scream.

5 0

2 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo

faltersainse [42]

Answer:

Length of pipe $= 0.057$ meter

Explanation:

Speed of a transverse wave on a string

$v = \sqrt{\frac{F}{\mu} }$

where F is the tension in string and $\mu$ is the mass per unit length

Thus,

$\mu = \frac{m}{L}$

Substituting the given values we get -

$\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}$

Speed of a transverse wave on a string

$v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}$

For third harmonic wave , frequency is equal to

$f = \frac{nv}{2L}$

Substituting the given values, we get -

$f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08$

Length of pipe

$L = \frac{nv}{4 f}$

Substituting the given values we get

$n = 1$ for first harmonic wave

$L = \frac{344* 1}{4*1502.08} \\L = 0.057$

Length of pipe $= 0.057$ meter

5 0

3 years ago