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2 years ago

Why is it difficult to define abnormal behavior?

Physics

1 answer:

Ivanshal [37]2 years ago

5 0

Answer: A. There is a strict definition of abnormal behavior based on an established set of social norms.

Forgive me if I’m wrong :C

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A hot air balloon contains 85,000 moles of air. To what temperature must the

Oksi-84 [34.3K]

Answer:

B is the best answer for this question

8 0

3 years ago

A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is

avanturin [10]

Answer: 22.6 hours

Explanation:

The power is the measure of the rate of energy.

In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as

$P=IV$

where

I is the current

V = 12.0 V is the voltage of the battery

So the energy delivered by the battery can be written as

$E=Pt=VIt$

Where

$It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s$

So the energy delivered is

$E=(12.0)(183,600)=2.2\cdot 10^6 J$

At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:

$t=\frac{2.2\cdot 10^6 J}{27.0 W}=81481 s = 22.6 h$

3 0

3 years ago

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.

6 0

3 years ago

A glass of juice at 20°C was kept in the open. After 30 minutes, the temperature of the Juice remained the same. Which statement

devlian [24]

The answer would be B the temperature of the juice was the temperature of the surrounding air

7 0

3 years ago

Read 2 more answers

How much time does it take a person to walk 12km north at the velocity at a velocity of 6.5 km/hrs?

alukav5142 [94]

"6.5 km/hr" is not a velocity. It's just a speed, so
we don't know what direction he's walking.

If he happens to be walking north, then it takes him

(12 km) / (6.5 km/hr) = 1.846... hours (rounded) .

If he's walking in any other direction, it takes him longer than that.

If the angle between north and the direction he's walking is
90 degrees or more, then he can never cover any northward
distance, no matter how long he walks.

6 0

3 years ago

Read 2 more answers