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3 years ago

Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss

Physics

1 answer:

vekshin13 years ago

4 0

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of transfer by

Explanation:

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Determine the mass of a boat that is propelled forward by a force of 1250N at an acceleration of 3 m/s2.

Pavlova-9 [17]

Answer:

416.6 KG

Explanation:

Due to the mass being missing from the equation you divide the force by the acceleration

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3 years ago

What is it called when objects become positively charged when they lose electrons and negatively charged when they gain electron

Alexeev081 [22]

It is c: electric charge

4 0

3 years ago

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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

4 years ago

A ball rolls up a ramp with a velocity of 8.0 m/s. How high up the ramp does it travel?

Flura [38]

The greatest height the ball will attain is 3.27 m

<h3>Data obtained from the question</h3>

Initial velocity (u) = 8 m/s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =?

The maximum height to which the ball can attain can be obtained as follow:

v² = u² – 2gh (since the ball is going against gravity)

0² = 8² – (2 × 9.8 × h)

0 = 64 – 19.6h

Collect like terms

0 – 64 = –19.6h

–64 = –19.6h

Divide both side by –19.6

h = –64 / –19.6h

h = 3.27 m

Thus, the greatest height the ball can attain is 3.27 m

Learn more about motion under gravity:

brainly.com/question/13914606

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3 years ago

Giving lots of points

bekas [8.4K]

Answer:

wind turbine

Explanation:

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3 years ago

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