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3 years ago

A forest ranger might be ranked as being high in what type of intelligence?

Physics

1 answer:

Drupady [299]3 years ago

7 0

<h2>Answer:</h2>

<u>A forest ranger might be ranked as being high in </u><u>Naturalistic</u><u> intelligence.</u>

<h2>Explanation:</h2>

Naturalistic intelligence is the type of intelligence that relates the activities of people that how sensitive an individual is to nature and the world. Zookeepers, biologists, gardeners, and veterinarians are among those people who have a higher capability of having high naturalist intelligence. Similarly a ranger who is fighting in forests must also be kept in the same category and he might be ranked as being high in the field of Naturalistic Intelligence.

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A body of mass 2 kg at O has an initial velocity of 3m/s along OE and it is subjected to a force of 4N perpendicular to OE the d

timofeeve [1]

Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²

Perpendicular distance:

s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m

Horizontal distance:
s = ut
= 3 x 4
= 12 m

Total distance = √(12² + 16²)
= 20 m.

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3 years ago

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2 years ago

Read 2 more answers

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known 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, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

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

where, $\vec{P_\lambda}$ 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,

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

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

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

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