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3 years ago

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Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rot

ates in a circular orbit about this nucleus. In the n = 1 orbit the electron is 5.29 10-11 m from the nucleus and it rotates with an angular speed of 4.12 1016 rad/s. Determine the electron's centripetal acceleration in m/s2.

1 answer:

riadik2000 [5.3K]3 years ago

7 0

To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as

$a_c = \omega^2 r$

Here,

$\omega =$ Angular velocity

r = Radius

Our values are given as,

$\omega = 4.12*10^{16}rad/s$

$r = 5.29*10^{-11}$

Replacing,

$a_c = (4.12*10^{16})^2( 5.29*10^{-11})$

$a_c = 8.979*10^{22}m/s^2$

Therefore the electron's centripetal acceleration is $8.979*10^{22}m/s^2$

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Answer:

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Explanation:

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V=(4/3)πr³ ⇒

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and r=14cm then

$\frac{dV}{dt}=4\pi(14cm)^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=4\pi196cm^{2}(-0.1\frac{cm}{min})\\\frac{dV}{dt}=-78,4\pi \frac{cm^{3} }{min}$

Note: as you can see the relationship of change of r with respect to t is negative because it is a decrease, and also its volume ratio

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2 years ago

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This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me

Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

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1 year ago

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3 years ago

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Minchanka [31]

Answer:

(-1.5,-1.5)m

Explanation:

we know that:

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where $X_{cm}$ is the location of the center of gravity in the axis x, $m_i$ is the mass of the object i and $x_i$ the first coordinate of center of gravity of object i.

so:

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Where $x_4$ is the first coordinate of the center of gravity for the fourth object.

Therefore, solving for $x_4$ , we get:

$x_4 = -1.5m$

At the same way:

$Y_{cm} = \frac{m_1y_1+m_2y_2....m_ny_n}{m_1+m_2...m_n}$

where $Y_{cm}$ is the location of the center of gravity in the axis y, $m_i$ is the mass of the object i and $y_i$ the second coordinate of center of gravity of object i. replacing values we get:

$Y_{cm} = \frac{(5kg)(0)+(3kg)(4)+(4kg)(0)+(8kg)y_4}{5+3+4+8}$

Where $y_4$ is the second coordinate of the center of gravity for the fourth object.

solving for $y_4$ :

$y_4 = -1.5m$

It means that the object of mass 8kg have to be placed in the

coordinates (-1.5,-1.5) m.

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3 years ago