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3 years ago

The current atomic model has a

2 answers:

8 0

he current atomic model has a predicted area surrounding the nucleus where the electron could be located thanks to quantum physics, subatomic particles makes leaps through space, and an electron could even be a million miles from its nucleus.

So the correct option among the other answer choices are as follows: C.

Oksanka [162]3 years ago

3 0

The current atomic model has a predicted area surrounding the nucleus where the electron could be located thanks to quantum physics, subatomic particles makes leaps through space, and an electron could even be a million miles from its nucleus.

So the correct option among the other answer choices are as follows: C.

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Does a five pound ball weigh more or less 40000 miles above sea level

Ann [662]

It would weigh less bc the higher you are the less gravity there is.

5 0

1 year ago

If a sprinter has an average acceleration of 9.5 m/s2, how long does it take for the person to go from rest to 10 m/s

mojhsa [17]

Answer:bruh Brandon I’m trying to get the answer I look it up on google and I see this I’m dead see you in 4th hour

Explanation:

8 0

2 years ago

A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

$\rho =$ Linear density

Our data are given by

Tension , T = 70 N

Linear density , $\rho = 0.7 kg/m$

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

$V = 10m/s$

Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

$f = \frac{1}{T}$

$f = \frac{1}{0.35}$

$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

3 0

3 years ago

It takes 5.0 seconds for a wave with a wavelength of 2.5 m to travel past your location. What’s the period of the wave. What is

Anna35 [415]

(a) Period of the wave
The period of a wave is the time needed for a complete cycle of the wave to pass through a certain point.
So, if an entire cycle of the wave passes through the given location in 5.0 seconds, this means that the period is equal to 5.0 s: T=5.0 s.

(b) Frequency of the wave
The frequency of a wave is defined as
$f= \frac{1}{T}$
since in our problem the period is $T=5.0 s$ , the frequency is
$f= \frac{1}{5.0 s}=0.2 Hz$

(c) Speed of the wave
The speed of a wave is given by the following relationship between frequency f and wavelength $\lambda$ :
$v=f \lambda = (0.2 Hz)(2.5 m)=0.5 m/s$

8 0

3 years ago

A light-year is the distance light travels in one year (at speed = 2.998 × 108 m/s). (a) how many meters are there in 11.0 light

larisa [96]

<span>The answers are as follows:

(a) how many meters are there in 11.0 light-years?

11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m

(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au

(c) what is the speed of light in au/h? au/h

</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h

8 0

2 years ago