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2 years ago

6

It is late and Carlos is sliding down a rope from his third-floor window to meet his friend Juan. As he slides down the rope fas

ter and faster, he becomes frightened and grabs harder on the rope, increasing the tension in the rope. As soon as the upward tension in the rope becomes equal to his weight:

1 answer:

gavmur [86]2 years ago

3 0

Answer:

It would help Carlos stop sliding down and he would stay in his place in the air.

Explanation:

According to Newton's 1st Law of Motion, a body is said to be at rest or uniform motion when no net force is acting on it. This explains that when the tension in the rope becomes equal to Carlos' weight, the net force becomes zero which means the body (Carlos), is at rest and remains in the air instead of sliding down the rope.

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7.For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isot

Oksana_A [137]

I believe the answer to this is Zinc

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3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

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3 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

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2 years ago

Which of the following best describes the human population from early times to the present?

Makovka662 [10]

Answer:

We have a not significant increase of the population until 1700s or 1800s and then a significant increase growth from these years to the present.

Explanation:

From the figure attached we see the evolution of the human population since early times (1050).

We see that from 1050 until 1750-1850 we have an increase slowly with a low value for the increase per year.

But after these years (1750-1850) we see a considerable increase of the population, like an exponential model.

So then we can conclude in general terms this:

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2 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

7 0

2 years ago