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Gala2k [10]
2 years ago
15

When you start from your house, your bike’s speedometer dial shows a reading of 842 Km. After you go on a straight road for half

an hour the dial shows 854km. At this point you pickup your friend and take a right turn and travel on a straight road for another 15 minutes. At the end your dial reads 859 km. Find the magnitude of your displacement?
Physics
1 answer:
ArbitrLikvidat [17]2 years ago
4 0

Answer:

S = 12757.5 m

Explanation:

A magnitude of the displacement can be obtained by visualizing the walking. The actual path from A to B Is 3 m then from B to D as 5 m and finally from D to E as 6 m. |S| =√92+52 = 10.29 m. The direction of Resultant displacement is South East.

<u>Solving for the different variables we can use the following formulas:  </u>

  • Given v, u, and t solve for s

Given initial velocity, final velocity, and time calculate the displacement.

s = ½( v + u )t

  • Given S, T, and U solve for V

Given displacement, time, and initial velocity calculate the final velocity.

v = 2s/t - u

  • Given S, T, and V solve for U

Given displacement, time, and final velocity calculate the initial velocity.

u = 2s/t - v

  • Given v, u, and t solve for s

Given initial velocity, final velocity, and time calculate the displacement.

t = 2s/ ( v + u )

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Approximately 5.11 \times 10^{-19}\; {\rm J}.

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

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Look up Planck's constant: h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}.

Apply the Rydberg formula to find the wavelength \lambda (in vacuum) of the photon in question:

\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}.

The frequency of that photon would be:

\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}.

Combine this expression with the Rydberg formula to find the frequency of this photon:

\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}.

Apply the Einstein-Planck equation to find the energy of this photon:

\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}.

(Rounded to three significant figures.)

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