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1 year ago

An athlete walks 8 meters East , 2 meters South, 8 meters West, and finally 2 meters North. What is the total distance traveled?

Physics

2 answers:

ser-zykov [4K]1 year ago

8 0

Answer:

the sum of all the values which is 20 meters

Explanation:

total is everything combined

serg [7]1 year ago

6 0

Explanation:

20 meters is travelled by him

please mark me as brainliest

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An atom of a certain element has 36 protons, 36 electrons, and a mass number of 84. At room temperature, this element is a very

Grace [21]

Answer:

12 neutrons

Explanation:

The number of protons also shows the atomic number. Therefore the element in question is Krypton (Kr), which also is a noble gas.

Neutrons = Mass number - protons - electrons

Here neutrons = 84 - 36 - 36 = 12

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

Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snork

fgiga [73]

Answer:

2354.4 Pa

40221 Pa

Explanation:

$\rho$ = Density = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth

The pressure difference would be

$\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa$

The pressure difference in the first case is 2354.4 Pa

$\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa$

The pressure difference in the second case is 40221 Pa

7 0

3 years ago

The pressure of air pushing down on a house roof is extremely large. Why is the roof not crushed?

Degger [83]

There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.

7 0

3 years ago

A cube of plastic 1.2x10^-5 km on a side has a mas of 1.1 g. what is its density in g/cm^3?

Mamont248 [21]

Density = mass / volume

mass = 1.1 g

volume = length of side ^ 3 = [1.2 * 10^-5 km * 100000 cm/km]^3 = [1.2 cm]^3 = 1.728 cm^3

density = 1.1 g / 1.728 cm^3 = 0.64 g / cm^3

5 0

3 years ago