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alexgriva [62]
3 years ago
14

In a hydrogen atom, the electron in orbit around the proton feels an attractive force of about 7.45 × 10−8 N. If the radius of t

he orbit is 5.89 × 10−11 m, what is the frequency? Answer in units of rev/s
Physics
1 answer:
Tcecarenko [31]3 years ago
8 0
The attraction force provides the electron's centripetal force. 

<span>8.30^-8N = mrω² </span>

<span>ω² = 8.30^-8 / (9.11^-31 kg x 4.70^-11m) .. .. ω² = 1.94^33 (rad/s)² .. .. ω = 4.40^16 rad/s </span>

<span>f = ω/2π = 4.40^16 / 2π .. .. .. ►f = 7.0^15 Hz</span>
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The ____ of a position-time graph represents an object’s velocity.
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Answer:

The slope of a position-time graph represents an object’s velocity.

Explanation:

In a position-time graph, the values on the x-axis represent the time, while the values on the y-axis represent the position of the object.

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v=\frac{\Delta s}{\Delta t}

However, we can see that this definition corresponds to the slope of the curve in a position-time graph. In fact:

\Delta s, the displacement, corresponds to the difference in position, so the difference between the values on the y-axis: \Delta s=y_2 -y_1

\Delta t, the time interval, corresponds to the difference in times, so the difference between the values on the x-axis: \Delta t= t_2 -t_1=x_2 -x_1

So, the velocity is

v=\frac{\Delta s}{\Delta t}=\frac{y_2 -y_1}{x_2 -x_1}

which corresponds to the slope of the curve.

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A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

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