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Dafna11 [192]
3 years ago
15

What is the ecliptic, and why is it tilted with respect to the celestial equator?

Physics
1 answer:
Dahasolnce [82]3 years ago
4 0
The ecliptic is the line in the celestial sphere traced out by the sun as the rotates (the path in the sky that the sun takes over head during one 24hr day). The celestial equator is just the extension of the regular equator on the surface of the earth.

This means that there will be a "tilt" between the ecliptic and the celestial equator that depends on your latitude of the observers location on the earths surface.

source: astrophysicist
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A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force
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Te direction of the magnetic force for the velocity of the proton in the

-ve  y direction will be +ve z direction.

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So,  we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x  direction,  so the magnetic force is acting in the positive z direction.

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A boy pushes forward a cart of groceries with a total mass of 40.0 kg. What is the acceleration of the cart if the net force on
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Read 2 more answers
A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let 'F_A' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

F_A × 10.0 m - W × 4.0 m = 0

∴ F_A × 10.0 m = W × 4.0 m = 600 N × 4.0 m

F_A × 10.0 m = 600 N × 4.0 m

∴  F_A = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', F_A = 240 N.

8 0
3 years ago
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