Which of the following best describes a chemical equation that demonstrates the law of conservation of matter?

1 answer:

3 0

B, since it is the only one that actually conserves matter for certain. In each of the others, matter could still be imbalanced, since for A, for example, it could be 5 Carbons on the right and 5 Chlorines on the left, and that would not balance.

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Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the

Leviafan [203]

Answer:

in first case the torque is maximum.

Explanation:

Torque is defined as the product of force and the perpendicular distance.

τ = F x d x Sinθ

In case A: the angle between force vector and the distance vector is 90 so torque is

τ = F x d

In case B: the angle between force vector and the distance is 30°.

τ = F x d x Sin30

τ = 0.5 Fd

So the torque is maximum in first case.

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sesenic [268]

Answer:

5.6ft

Explanation:

on avgerage a male gets 5 foot 6 inches

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A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,