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2 years ago

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a driver is going at 120km h and sees a barrier 60.0 m ahead it takes 5secounds to apply the brakes and decelerates at 12m does

the driver hot the barrier

1 answer:

Soloha48 [4]2 years ago

5 0

Answer:

yes, if you're going at 120 km and you saw the wall that late then it wouldn't me possible to decrease 12 meters in 5 seconds and not hit the wall that's only 60 meters away

Explanation:

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what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three

grin007 [14]

Answer:

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.

Look up the Rydberg constant for hydrogen: $R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}$ .

Look up the speed of light in vacuum: $c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}$ .

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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1 year ago

What happens when a star's core runs out of hydrogen and why does this occur?

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2 years ago

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What does the principal quantum number determine?

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2 years ago

What is the purpose of chloroplasts, and why do you think it is missing from the root hair cell?

Karolina [17]

Explanation:

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2 years ago

Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu

anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

$\frac{P_{1}}{9810}$ + $\frac{V_{1} ^{2}}{19.62}$ + $h_{1}$ = $\frac{P_{2}}{9810}$ + $\frac{V_{2} ^{2}}{19.62}$ + $h_{2}$ --------------(1)

Where

$P_{1}$ = Gauge pressure at inlet = 3.70105 pascal

$V_{1}$ = velocity at inlet = 2.4 $\frac{m}{sec}$

$P_{2}$ = Gauge pressure at outlet = we have to calculate

$V_{2}$ = velocity at outlet = 3.5 $\frac{m}{sec}$

$h_{2} - h_{1}$ = 3.6 m

Put all the values in equation (1) we get,

⇒ $\frac{3.70105}{9810}$ + $\frac{2.4 ^{2}}{19.62}$ = $\frac{P_{2}}{9810}$ + $\frac{3.5 ^{2}}{19.62}$ + 3.6

⇒ 0.294 = $\frac{P_{2}}{9810}$ + 0.6244 + 3.6

⇒ $\frac{P_{2}}{9810}$ = 0.294 - 0.6244 - 3.6

⇒ $\frac{P_{2}}{9810}$ = - 3.9304

⇒ $P_{2}$ = - 38557.224 pascal

This is the value of gauge pressure at outlet.

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