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4 years ago

Why hydrogen atom does not emit X-rays??

Physics

1 answer:

goblinko [34]4 years ago

8 0

b) it has narrow energy levels is your answer.

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Recalculate 100 km/h to m/s

qwelly [4]

= 27.777

Explanation:

A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.

4 0

3 years ago

A 50-loop circular coil has a radius of 3 cm. It is oriented so that the field lines of a magnetic field are perpendicular to th

nikitadnepr [17]

Answer:

-17.8 V

Explanation:

The induced emf in a coil is given as:

$E = \frac{-NdB\pi r^2}{dt}$

where N = number of loops

dB = change in magnetic field

r = radius of coil

dt = elapsed time

From the question:

N = 50

dB = final magnetic field - initial magnetic field

dB = 0.35 - 0.10 = 0.25 T

r = 3 cm

dt = 2 ms = 0.002 secs

Therefore, the induced emf is:

$E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V$

Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.

6 0

3 years ago

In what part of the Milky Way galaxy are the stars packed closest together?

valentina_108 [34]

Answer: the nucleus

Explanation: The central nucleus is the oldest and most densely packed part of the milky way.

6 0

3 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

8 0

3 years ago

Which of the following is NOT available for drinking

Effectus [21]

Answer:

Explanation:

lakes are too big or large

4 0

3 years ago