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Serhud [2]
2 years ago
14

Which statement describes one way in which global winds affect weather and climate? A. Polar easterlies move warm air to the mid

latitudes B. Global winds cause deep ocean currents to form C. In the northern hemisphere, the winds spiral to the left D. The jet stream carries warm air to the north
Physics
2 answers:
Genrish500 [490]2 years ago
4 0

The answer your looking for is "D".

nydimaria [60]2 years ago
4 0

Polar easterlies move cool air to the mid latitudes

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A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle
alexira [117]
<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
  • According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
  • Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

       = 0.3759

r = Sin^-1 0.3759

   = 22.08

   <u>≈ 22°</u>

3 0
2 years ago
charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁  + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³  N / C .

3 0
2 years ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
3 0
2 years ago
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its
Alja [10]

Answer:

 Terminal velocity of object = 12.58 m/s

Explanation:

 We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = 12.0v+4.00v^2

Equating both, we have

    784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0

  So v = 12.58 m/s or v = -15.58 m/s ( not possible)

 So terminal velocity of object = 12.58 m/s    

7 0
2 years ago
Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part
erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

<em>x= 2 cm to the left of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

\displaystyle f=k\frac{q_1\ q_2}{d^2}

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}

\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}

\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}

\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}

\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}

Equating

\displaystyle F_{13}=F_{23}

\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}

Operating and simplifying

\displaystyle (0.02-x)^2=4x^2

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

\displaystyle 0.02-x=\pm 2x

Assuming the positive sign :

\displaystyle 0.02-x= 2x

\displaystyle 3x=0.02

\displaystyle x=0.00667\ m

x=0.67\ cm

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

\displaystyle 0.02-x=-2x

\displaystyle x=-0.02\ m

\displaystyle x=-2\ cm

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

5 0
2 years ago
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