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2 years ago

Question 3

Physics

1 answer:

pickupchik [31]2 years ago

4 0

When a police officer is trying to decide if a driver is speeding, what is his point of reference. The speed limit

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You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s

Jobisdone [24]

Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
Since the speed of sound and light remains constant in a particular medium, we can use
Speed = Distance/Time

For light,
3 x 10^8 = d/t
t = d/(3 x 10^8) -1

For sound,
339 = d/(t + 30) -2

Putting value from 1 in 2.
d = 10^4 m(approx)

3 0

3 years ago

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.)

6 0

2 years ago

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x

Andrei [34K]

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$

6 0

3 years ago

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Naya [18.7K]

Answer:

$2.966\times 10^{-11}\ N$

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = $6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}$

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

$F=\dfrac{GMm}{d^2}$

Plug in the given values and solve for 'F'. This gives,

$F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})$

The same force is experienced by both cannonball and marble.

Therefore, the gravitational force of the marble is $2.966\times 10^{-11}\ N$

3 0

2 years ago

Which factors affect the gravitational force between objects? Check all that apply

neonofarm [45]

Answer:distance between objects and the masses of the objects

Explanation:

Gravitational force of attraction between two masses is directly proportional to the product of the masses and inversely proportional to the square of distance apart

5 0

2 years ago