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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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A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?
yanalaym [24]

Answer:

KE=12,500J

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KE = \frac{1}{2}mv^2

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7 0
2 years ago
A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
blondinia [14]

Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

8 0
3 years ago
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