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Inga [223]
1 year ago
7

The gravitational force, Fbetween an object and the Earth is inversely proportional to the square of the distance from the objec

t to the center of the Earth. If an astronaut weighs 214 pounds on the surface of the Earth, what will this astronaut weigh 500 miles above the Earth? Assume that the radius of the Earth is 4000m / l * es Round off your answer to the nearest pound)

Physics
1 answer:
yulyashka [42]1 year ago
7 0

ANSWER:

169 pounds

STEP-BY-STEP EXPLANATION:

We can calculate the weight of the astronaut since we know the relationship between both variables (weight and radius), so we establish the following proportion:

\frac{W_2}{W_1}=\frac{R^2}{(R+d)^2}

The astronaut's initial weight is 214 pounds (W1), the radius is 4000 miles (R) and the distance is equal to 500 miles, we substitute and calculate the new weight of the astronaut as follows:

\begin{gathered} W_2=W_1\cdot\frac{R^{2}}{(R+d)^{2}} \\  \\ \text{ We replacing} \\  \\ W_2=214\cdot\frac{4000^2}{\left(4000+500\right)^2} \\  \\ W_2=169.086=169\text{ pounds} \end{gathered}

The astronaut's new weight is 169 pounds

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The focal length of david's lens is 60 cm. if rebecca stands in front of david at a distance of do and david perceives the posit
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if rebecca stands in front of david at a distance of do and david perceives the position of rebecca at di, di will be +84 cm

<h3>What is focal length ?</h3>

How strongly light converges or diverges depends on an optical system's focal length, which is the inverse of optical power. A system with a positive focus length is said to converge light, whereas one with a negative focal length is said to diverge light.

focal length = +60 cm

magnification m = -0.40

focal length being positive an magnification negative.

given lens is a convex lens.

for a lens

m = di/do and 1/f = (1/di) - (1/do)di

= -0.4do1/f = (1/-0.4do) - 1/do do

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Edwin Hubble calculated the expansion rate of the universe. On what evidence did he base his calculation?
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Edwin Hubble calculated the expansion rate of the universe. The evidence that he base his calculation is the differences in redshift for galaxies. The answer is letter B. the red shift of galaxies was directly proportional to the distance of the galaxy from earth.  It means that bodies farther away from Earth were moving away faster. The Hubble’s constant is the ratio of distance to redshift equal to 170 kilometers per second per light year of distance.

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An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's e
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Answer:

a) \omega = 10.407\,\frac{rad}{s}, b) m = 4.617\,kg, c) A = 1.355\,m

Explanation:

a) The system have a simple armonic motion, whose position function is:

x(t) = A\cdot \cos (\omega\cdot t + \phi)

The velocity function is determined by deriving the position function in terms of time:

v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)

The acceleration function is found by deriving again:

a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)

Let assume that t = 0\,s. The following nonlinear system is built:

A\cdot \cos \phi = 0.660\,m

-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}

-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}

System can be reduced by divinding the second and third expressions by the first expression:

\omega \cdot \tan \phi = 18.636\,\frac{1}{s}

\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}

Now, the last expression is divided by the first one:

\omega = 10.407\,\frac{rad}{s}

b) The mass of the block is:

m = \frac{k}{\omega^{2}}

m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }

m = 4.617\,kg

c) The phase angle is:

\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega}  \right)

\phi \approx 0.338\pi

The amplitude is:

A = \frac{0.660\,m}{\cos 0.338\pi}

A = 1.355\,m

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