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3 years ago

PLZZ HELP ME PLZ I NEED YALL HELP WITH THIS LAST QUESTION

Physics

1 answer:

Mamont248 [21]3 years ago

6 0

Answer:

The 5 Forces

Explanation:

The five forces that influence wind speed and direction are: Pressure gradient force (flow from high to low pressure) Coriolis force (apparent deflecting force due to the rotation of the Earth) Turbulent drag (Earth's surface or objects like trees or grass resist air flow and decrease wind speed near the ground)

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3 years ago

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A girl starts from rest and reaches a walking speed of 1.4 m/s in 3.0 s. She walks at this speed for 6.0 s. The girl then slows

Alona [7]

Answer:

1) The girl's acceleration at time 't' is, $a=0.47$ m/s²

2) The girl's acceleration at time 't₁' is, $a_{1}=0$ m/s²

3) The girl's acceleration at time 't₂' is, $a_{2}=-0.14$ m/s²

Explanation:

Given data,

The initial walking speed of the girl, u = 0

The speed of the girl at the time 't' 3 s is, v₁ = 1.4 m/s

The time period the girl walked at the speed 1.4 m/s is, t₁ = 6 s

The girl slows down and comes to a stop during a period, t₂ = 10 s

1) The girl's acceleration at time 't'

$a=\frac{v-u}{t}$

$a=\frac{1.4-0}{3}$

$a=0.47$ m/s²

2) The girl's acceleration at time 't₁'

$a_{1} =\frac{v_{1} -v}{t_{1}}$

$a_{1}=\frac{1.4-1.4}{6}$

$a_{1}=0$ m/s²

3) The girl's acceleration at time 't₂'

$a_{2} =\frac{v_{2} -v_{1}}{t_{2}}$

$a_{2}=\frac{0-1.4}{10}$

$a_{2}=-0.14$ m/s²

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3 years ago

The surface temperature of a planet depends on both the distance of the planet from the Sun and on how the panet's atmosphere di

BaLLatris [955]

Answer:

Aphelion: 6404 W/m2

Perihelion: 14978 W/m2

Explanation:

The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.

$\Phi sol = \frac{Psol}{4 * \pi * d^2}$

The distances we need are the aphelion and perihelion of Mercury.

Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:

$e = \frac{c}{a}$

Where

e: eccentricity

c: linear eccentricity

a: semi major axis

The linear eccentricity is equal to the distance of the focus of the center of the ellipse.

$c = a * c =$

a = 0.39 AU = 5.83e10 m

$c = 5.83e10e * 0.21 = 1.22e10 m$

In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:

Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m

Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m

And the solar energy fluxes will be:

$\Phi Ap = \frac{4e26}{4 * \pi * 7.05e10^2} = 6404 W/m2$

$\Phi Pe = \frac{4e26}{4 * \pi * 4.61e10^2} = 14978 W/m2$

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3 years ago

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. Which of the velocity components (n

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Green because it is the color it does not absorb.
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