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3 years ago

How was Bohr's atomic model different from those of previous scientists?

1 answer:

4 0

Answer:Bohr placed the electrons in distinct energy levels. Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits.

Explanation: also rutherfords was just a hypothesis while Bhor took the time to make his an experiment

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Which special effects technique is being used in television weather reports in which meteorologists stand in front of moving map

Pachacha [2.7K]

A. cgi... they're usually filmed with a greenscreen and is keyed out in editing. known as chroma keying

6 0

3 years ago

Read 2 more answers

What is the momentum of an object with 5.22 kg and 1.7 m/s

BlackZzzverrR [31]

Answer:

8.874

Explanation:

You need to times 5.22 kg and 1.7 m/s to get 8.874.

8 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

In his motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing spot, a distance of 120120 mi, in 33 hr.

photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

Explanation:

x = the rate of the boat in still water

y = the rate of the current.

Distance travelled = 120 mi

Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

$\frac{120}{3}=x-y\\\Rightarrow 40=x-y$

Speed downstream

$\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y$

Adding both the equations

$48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x$

$40=44-y\\\Rightarrow 40-44=-y\\\Rightarrow y=4$

The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>

8 0

3 years ago

You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in

Vlad1618 [11]

Answer:

$0.28 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

$u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s$ is the initial velocity

$v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s$ is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

$a=\frac{2.78-2.22}{2}=0.28 m/s^2$

5 0

3 years ago