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3 years ago

Winds near the surface in the continental United States generally flow from west to east. What causes this prevailing wind?

Physics

1 answer:

NNADVOKAT [17]3 years ago

8 0

I want to say that the answer is either A or B but im not too sure. Im leaning more towards A tho

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I E I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I E I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or

Ostrovityanka [42]

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

$r' = \frac{r}{2}$

So the new force will be

$F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F$

So, the force will quadruple.

4 0

3 years ago

the three major scales used to measure earthquakes are Mercalli scale Richard scale and magnitude scale

iVinArrow [24]

False; the three major scales used to measure earthquakes are the Mercalli Scale, the Richter Scale, and the Magnitude Scale. I hope this helps!

3 0

3 years ago

Hi, Solve for λ E=hc/λ

Paul [167]

Answer:

λ=hc/E

Explanation:

E=hc/λ

Eλ=hc

λ=hc/E

4 0

3 years ago

How do we represent Earth’s face?

jeka94

Answer:

Human-driven changes in arrive utilize and arrive cover such as deforestation, urbanization, and shifts in vegetation designs moreover change the climate, coming about in changes to the reflectivity of the Soil surface (albedo), emanations from burning timberlands, urban warm island impacts and changes within the normal water cycle.

6 0

3 years ago