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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

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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Suppose you have a 113-kg wooden crate resting on a wood floor. (μk = 0.3 and μs = 0.5) (a) What maximum force (in N) can you ex

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3 years ago

Set local ground level to 700 ft, and record the inches of mercury.

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2 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

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Given:

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angle of projectile from the horizontal, $\theta=20^{\circ}$

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$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

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$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

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<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

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$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

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3 years ago

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Dmitry [639]

I think the word is "baton"

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4 years ago

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