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3 years ago

A Stegosaurus hits' a wall with a force of

Physics

1 answer:

Pavel [41]3 years ago

3 0

Answer: a= -1.9 m/s²

Explanation: F = ma and a = F / m = 60 N / 32 kg = 1.875 m/s²

Because of law of action and reaction, wall applies same force to dino

In opposite direction and Dino is decelerating. Thus sign Is negative.

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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2

Nookie1986 [14]

<h2>Answers:</h2>

According to Newton's Law of Gravitation, the Gravity Force is:

$F=\frac{GMm}{{r}^{2}}$ (1)

This expression can also be written as:

$F=GMm{r}^{-2}$ (2)

If we derive this force $F$ respect to the distance $r$ between the two masses:

$\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr$ (3)

Taking into account $GMm$ are constants:

$\frac{dF}{dr}dFdr=-2GMm{r}^{-3}$ (4)

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies. </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction. </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit $N/km$ :

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

We have a force that decreases with a rate 1 $\frac{dF_{1}}{dr}dFdr=4N/km$ when $r=20000km$ :

$4N/km=-2\frac{GMm}{{(20000km)}^{3}}$ (6)

Isolating $-2GMm$ :

$-2GMm=(4N/km)({(20000km)}^{3})$ (7)

In addition, we have another force that decreases with a rate 2 $\frac{dF_{2}}{dr}dFdr=X$ when $r=10000km$ :

$XN/km=-2\frac{GMm}{{(10000km)}^{3}}$ (8)

Isolating $-2GMm$ :

$-2GMm=X({(10000km)}^{3})$ (9)

Making (7)=(9):

$(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}$ (10)

Then isolating $X$ :

$X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}$

Solving and taking into account the units, we finally have:

$X=-32N/km$ >>>>This is how fast this force changes when $r=10000 km$

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4 years ago

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A student tested a variety of interacting forces to determine how they would result in motion of an object. If the student used

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D) The variable shown by letter C would result in a movement of the object to the right.

Explanation:

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3 years ago

The energy that a substance contains is

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The energy that substance contain is. B.heat

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A 4.3-g object moving to the right at 22 cm/s makes an elastic head-on collision with a 8.6-g object that is initially at rest.

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3 years ago

Light described as what two things?

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