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2 years ago

A nurse counts 66 heartbeats in one minute. What is the period of the hearts oscillation? In minutes

Physics

1 answer:

inn [45]2 years ago

7 0

The frequency is 66 / minute .

The period is 1/frequency = minute / 66 =

(60 seconds) / 66 = 0.91 second

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How do quantum numbers relate to electrons? please explain?

alukav5142 [94]

Quantum numbers allow us to both simplify and dig deeper into electron configurations. Electron configurations allow us to identify energy level, subshell, and the number of electrons in those locations. If you choose to go a bit further, you can also add in x,y, or z subscripts to describe the exact orbital of those subshells (for example 2px). Simply put, electron configurations are more focused on location of electrons then anything else.

Quantum numbers allow us to dig deeper into the electron configurations by allowing us to focus on electrons' quantum nature. This includes such properties as principle energy (size) (n), magnitude of angular momentum (shape) (l), orientation in space (m), and the spinning nature of the electron. In terms of connecting quantum numbers back to electron configurations, n is related to the energy level, l is related to the subshell, m is related to the orbital, and s is due to Pauli Exclusion Principle.

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2 years ago

A girl travels 20 miles on her bike, the trip takes two hours. express her speed in miles/hr

xxTIMURxx [149]

Answer:

10 mph

Explanation:

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2 years ago

If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:

$\theta = 1.22\frac{\lambda}{d}$

Where,

$\lambda$ = Wavelength

d = Width of the slit

$\theta$ = Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

$r = l\theta$

Relacing $\theta$

$r = l(\frac{1.22\lambda}{d})$

$r = 1.22\frac{l\lambda}{d}$

Replacing with our values we have that,

$r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})$

$r = 3680.91m$

Therefore the diameter of the beam on the moon is

$d = 2r$

$d = 2 * (3690.91)$

$d = 7361.8285m$

Hence, the diameter of the beam when it reaches the moon is 7361.82m

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3 years ago

What's earths most precious element?

atroni [7]

I believe the answer is californium because it is used in most metals and is very strong and expensive like gold and silver

3 0

3 years ago

Read 2 more answers

The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie

pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

$\bf \longrightarrow {v}^{2} = {u}^{2} + 2as \\ \\ \rm \longrightarrow {0}^{2} = {15}^{2} + 2 \times a \times 63 \\ \\ \rm \longrightarrow 0 = 225 + 126a \\ \\ \rm \longrightarrow 126a = - 225 \\ \\ \rm \longrightarrow a = - \dfrac{225}{126} \\ \\ \rm \longrightarrow a = - 1.78 \: m {s}^{ - 2}$

$\therefore$ Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0

2 years ago