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Nezavi [6.7K]
3 years ago
6

You are driving your car at a speed of 60 miles per hour, when you cross into Canada. Canada measures their speed in kilometers

per hour, and you need to figure out how fast you are going!​
Physics
1 answer:
Flauer [41]3 years ago
8 0

Answer:

Answer:

I don't know

Explanation:

I don't know

Explanation:

Answer:

I don't know

Explanation:

I don't know

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In a hydroelectric dam the potential energy of lake water is transformed into _ energy as it begins to more.
Marianna [84]

Answer:

mechanical energy

Explanation:

7 0
3 years ago
Read 2 more answers
Canada is the country (dash)of the united States
lana66690 [7]

I believe it is north.

5 0
3 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m/s. After 1.1 minutes, the finch ti
Romashka [77]

Answer:

Average Speed = 6.37 m/s

Explanation:

The average speed is simply given by the following formula:

Average Speed = Total Distance Traveled/Total Time Spent

here,

Total Time Spent = 1.1 min + 1.5 min = (2.6 min)(60 s/min) = 156 s

Now, for total distance, we have to calculate the distance traveled on tortoise and distance traveled while flying, separately. Therefore,

Distance Traveled on Tortoise = (Time spent on Tortoise)(Speed of Tortoise)

Distance Traveled on Tortoise = (1.1 min)(60 s/min)(0.06 m/s) = 3.96 m

Similarly,

Flying Distance = (Flying Time)(Flying Speed) = (1.5 min)(60 s/min)(11 m/s)

Flying Distance =  990 m

Since, total distance is the sum of both distances, therefore,

Total Distance = 3.96 m + 990 m = 993.96 m

Now, using the values in equation of average speed, we get:

Average Speed = 993.96 m/156 s

<u>Average Speed = 6.37 m/s</u>

4 0
3 years ago
Which is the correct definition for sea floor spreading
jok3333 [9.3K]
Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other.
6 0
3 years ago
Read 2 more answers
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