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3 years ago

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You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and tr

avels upward with a speed that varies with time according to v(t) = (3.0 m/s2)t + (0.20 m/s3)t2. When t = 4.0 s, what is the reading on the bathroom scale?

1 answer:

Genrish500 [490]3 years ago

5 0

Answer:

= 922N

Explanation:

m = 64kg

v(t) = (3.0 m/s²)t + (0.20 m/s³)t².

t = 4.0s

differentiation of v(t) so the acceleration is given by

$a(t) = \frac{ dv(t)}{dt} \\ = 3.0 + 0.4t\\ t = 4s\\a(4) = 3.0 + 0.4(4)\\= 4.6ms^-^2$

using Newton’s second law

F (net) = ΣFg = ma

⇒R - w = ma

R = m(a + g)

= 64 (9.81 + 4.6)

= 922N

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Answer:

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(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

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<h2>Answers:</h2>

<h2>(a) </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)

Or

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

<h2>(d) </h2>

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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