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3 years ago

Does the plane of the ecliptic coincide with the plane of the equator? explain.

1 answer:

3 0

<span>no it doesn't. one way to think of it is that the tropic of capricorn is the furthest south the sun's path goes, and the tropic of cancer is the northern most line that the sun traces out. since neither the tropic of capriforn nor the tropic of cancer are on the equator, the sun must be tracing out a different line and hence be on a different plane. also, if the sun was on the same plane as the equator, then we wouldn't have seasons.</span>

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PLEASE ANSWER ASAP

Sedbober [7]

Answer:

Explanation:

The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.

3 0

3 years ago

Rowenda is stopped on the side of the road. She then starts pedaling her bicycle for 15 seconds. Her final

lukranit [14]

Explanation:

acceleration = change in velocity / change in time

a = v2-v1

- - - - - -

V1 = 0 ( since she stopped)

V2 = 9 m/s

t =15s

a = 9 - 0

- - - -

= 0.6m/s^2

5 0

3 years ago

An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached

MakcuM [25]

Answer:

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

5 0

3 years ago

If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53Â° to the positive direction of the x axis, then what are (a) the x comp

Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a) Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

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6 0

2 years ago

draw a velocity graph with Vi = 4 m/s and decreasing uniformly so that velocity at 2 seconds is 2 m/s and remaining constant fro

Shkiper50 [21]

For the velocity graph: start at 0s and 4m/s and draw a straight line to 2s and 2 m/s. Then draw a straight horizontal line to 4s and 2m/s

For the acceleration graph: start with a horizontal line from 0s and 2m/s/s to 2s and 2m/s/s. The draw another line from 2 s and 0m/s/s to 4 s and 0m/s/s

6 0

3 years ago