Physics Homework Help

[cp]

1a)Calculate the period of a ball tied to a string of length 1.6m making 5.2 revolutions every second. Answer in units of s.

b) Calculate the speed of the ball. Answer in units of m/s.

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In a hydrogen atom, the electron in orbit around the proton feels an attractive force of about 7.67 x 10^-8 n. if the radius of the orbit is 5.72 x 10^-11 m what is the frequency in revolutions per second? Answer in units of rev/s.

-Pen

 

[Nullware]

1a)Calculate the period of a ball tied to a string of length 1.6m making 5.2 revolutions every second. Answer in units of s.

b) Calculate the speed of the ball. Answer in units of m/s.

--

In a hydrogen atom, the electron in orbit around the proton feels an attractive force of about 7.67 x 10^-8 n. if the radius of the orbit is 5.72 x 10^-11 m what is the frequency in revolutions per second? Answer in units of rev/s.

-Pen

1.
a) Period (P), Frequency (F)
P = 1 / F
P = 1 / 5.2
P = 0.1923s

The period is 0.1923s.

b) Circumference (C), Diameter (D), Radius (R), Distance (S), Speed (V), Laps (L), Time (T)
C = PI * D
C = PI * (R * 2)
C = PI * (1.6 * 2)
C = 10.053m

S = L * C
S = 5.2 * 10.053 = 52.2756m

V = S / T
V = 52.2756m / 1s = 52.28m/s

3.
Known equation: F = qE = m(v^2/r)
Electron mass: 9.1094*10^-31

Rewrite equation with what we have and need to find: F = m(v^2/r)
Rearrange equation: v = Sqrt(F*r/m)
v = Sqrt(7.67*10^-8 * 5.72*10^-11 / 9.1094*10^-31
v = 2194576.93 m/s

Revolution distance is the circumference
C = PI * (2 * r)
C = PI * (2 * 5.72*10^-11)
C = 3.594*10^-10 m

Revolutions per second is speed divided by total distance of one revolution
Frequency = v / C
F = 2194576.93 / 3.594*10^-10
F = 6.106*10^15 revs/s

The electron makes 6.106*10^15 revolutions per second.

 

[Alteranz]

f=1/T

therefore frequency = 1/period

 

[cp]

(12:04:56 AM) : can you thank viper and nullware for me in that thread

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