Ask question

Ask question

All categories

3 years ago

10

who described atoms as having a positive nucleus with electrons that have different energies at different distances from the nuc

leus.

1 answer:

zysi [14]3 years ago

4 0

Answer to your question : Niels Bohr

You might be interested in

???????????????????????????????????

anastassius [24]

What happened the thing is kinda glitched

8 0

2 years ago

What is the linear diameter (in meters) of an object that has an angular diameter of 10 arcseconds and a distance of 50,000 mete

PIT_PIT [208]

####### ##### ######

Explanation:

######### ###### #### #####

8 0

4 years ago

Read 2 more answers

Pls help me with this question

ryzh [129]

Answer:

16 ms2 is the answer for this question

8 0

4 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

4 years ago

Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observ

zhenek [66]

Answer:

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is: <u> </u>

From above; the distance between speaker A and speaker B can be expressed as:

$\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m$

The path length difference will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length will be half the wavelength; which is

= $\dfrac{1}{2}*4 \ m$

= 2 m

8 0

4 years ago