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3 years ago

The bohr model of the atom addressed the problem of

1 answer:

7 0

Rutherford's nuclear model stated that a cloud of negative electrons surround the nucleus however scientists realized that electrons in a cloud around the nucleus of an atom would be attracted to the nucleus, causing the atom to collapse. Thus Bohr's model proposed that electrons were contained in shells and they orbit the nucleus at fixed distances.

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______ can occur when water-saturated soil turns from a solid to a liquid as a result of an earthquake.

jek_recluse [69]

The answer is liquefaction

6 0

2 years ago

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele

evablogger [386]

Answer:

a)n= 3.125 x $10^{19$ electrons.

b)J= 1.515 x $10^{6$ A/m²

c) $V_{d$ =1.114 x $10^{4$ m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x $10^{-3$ m

radius 'r' = d/2 => 1.025 x $10^{-3$ m

no. of electrons 'n'= 8.5 x $10^{28}$

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x $10^{-19$ C

n= Q/e => 5/ 1.6 x $10^{-19$

n= 3.125 x $10^{19$ electrons.

b) the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x $10^{-3$ )²)

J= 1.515 x $10^{6$ A/m²

c) The typical speed' $V_{d$ ' of an electron is given by:

$V_{d$ = $\frac{J}{n|q|}$

=1.515 x $10^{6$ / 8.5 x $10^{28}$ x |-1.6 x $10^{-19$ |

$V_{d$ =1.114 x $10^{4$ m/s

d) According to these equations,

J= I/A

$V_{d$ = $\frac{J}{n|q|}$ = $\frac{I}{nA|q|}$

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

5 0

2 years ago

Read 2 more answers

According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s

Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

$\tau_{net} = 0$

now we have

$\frac{dL}{dt}= 0$

now we also know that

$Area = \frac{1}{2}r^2d\theta$

so rate of change in area is given as

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}$

so we will have

$\frac{dA}{dt} = \frac{1}{2}r^2\omega$

$\frac{dA}{dt} = \frac{L}{2m}$

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0

3 years ago

What is the equivalent resistance for a series circuit with three resistors : 5.0 ohms, 2.0 ohms, and 12.0 ohms

vesna_86 [32]

Answer:19ohms

Explanation:

equivalent resistance=5+2+12

equivalent resistance=19ohms

8 0

2 years ago

Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of

WINSTONCH [101]

a₀). You know ...
       -- the object is dropped from 5 meters
   above the pavement;
         -- it falls for 0.83 second.

a₁). Without being told, you assume ...
   -- there is no air anyplace where the marshmallow travels,
   so it free-falls, with no air resistance;
   -- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .

b). You need to find how much LESS than 5 meters
   the marshmallow falls in 0.83 second.

c). You can use whatever equations you like.
   I'm going to use the equation for the distance an object falls in
   ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d). To see how this all goes together for the solution, keep reading:

The distance that an object falls in ' T ' seconds
when it's dropped from rest is

   (1/2 G) x (T²) .

On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall

   (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

   (5.00 - 3.38) = 1.62 meters

above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0

2 years ago