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AlexFokin [52]
2 years ago
12

Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for

the four days of her trip. What kinds of conditions can she expect?
Physics
2 answers:
shepuryov [24]2 years ago
6 0
<h2>Answer:</h2>

Dry weather and mostly clear skies with larger diurnal temperature changes

<h2>Explanation:</h2>

High pressure is generally associated with nice weather having dry air that generally brings fair weather and light winds. When viewed from above, winds spiral out of a high-pressure center in a clockwise rotation in the Northern Hemisphere.  A warm front is associated with a high-pressure system and occurs when a warm air mass replaces a colder air mass therefore chances of rain are very less. So it is the best weather condition to travel.  

Zanzabum2 years ago
5 0

Calm, sunny days with wind moving away from the center.

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Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys
max2010maxim [7]

Answer:

The spring was compressed the following amount:

\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount \Delta x), and the total final energy is the addition of the kinetic energies of both masses:

E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2

E_i=E_f\\

\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

8 0
2 years ago
What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a
Fiesta28 [93]

Answer:

a) E_photon =0.306 eV

b) E_photon =0.166 eV

Explanation:

The energy of the photon (E) for n^th orbit of the hydrogen atom is given as:

E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})

where,

E_o = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})

E_photon =0.306 eV

b) Now for the transition from n = 5 to n = 6

E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})

E_photon =0.166 eV

7 0
3 years ago
Helppp pliss.............
Serggg [28]

i don't know sorry

free to comment if you know the answer

8 0
3 years ago
A 2 300-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.00 m before coming into contact
Mashcka [7]

Answer:

433903.8 N

Explanation:

From work energy theorem, the total work done is equivalent to change in kinetic energy. Kinetic energy is depedant on speed and since pile drive finally comes to rest, then final velocity is zero. Also, its initial velocity is not given. Therefore, the sum of work due to gravity and beam equals to zero. Work due to gravity is product of mass of pile driver, acceleration due to gravity and height while work due to beam is a product of force and distance. Substituting the given values then

2300*9.81*3+(F*0.156)=0

F=-433903.84615384615384615384615384615384615 N

Approximately, the magnitude of force is 433903.8 N and it acts upwards

6 0
3 years ago
Find the resultant vector (mag. and dir.) given the following information: Ax = 5.7, Ay = 3.4
Mashcka [7]

<u>Answer:</u>

  Magnitude = 6.64

  Direction = 30.82^0 from horizontal.

<u>Explanation:</u>

 We have horizontal component of vector Ax = 5.7 and vertical component of vector Ay = 3.4.

 Magnitude of vector acting perpendicularly = \sqrt{A_x^2+A_y^2}

 Substituting

    A=\sqrt{A_x^2+A_y^2} =\sqrt{5.7^2+3.4^2} =6.64

Direction θ = tan^{-1}(\frac{A_y}{A_x} )

  Substituting

      θ =tan^{-1}(\frac{A_y}{A_x} )= tan^{-1}(\frac{3.4}{5.7})=30.82^0

7 0
3 years ago
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