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3 years ago

Pls help ASAP!! Thank you

Physics

2 answers:

olga55 [171]3 years ago

7 0

Answer:

A. The south pole of one magnet is near the south pole of the other magnet.

Explanation:

Like poles repel.

Unlike poles attract.

Thus:

A should repel each other. (S and S together)

B might not be able to have enough attraction between both magnets to move towards each other so they remain still.

C should attract each other. (N and S together)

D might not be able to have enough repelling power between both magnets to repel each other so they remain still.

Overall, B and D would not have any attraction between each other so they are neutral. C is wrong because they have the strongest attraction while A has the strongest repelling in that situation.

GaryK [48]3 years ago

5 0

Answer:

Explanation:

The answer you should pick is A. Two magnets will attract each other when their poles are opposite and they are close together.

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2 years ago

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3 years ago

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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)

$g = G \cdot M/ r^{2}$ (equation 4)

where the universal gravitation constant $G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}$

Thus

$\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}$

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3 years ago