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DerKrebs [107]
3 years ago
8

Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what

is the magnitude of the gravitational force fg between the objects if the distance between them is 150 km ? express your answer in newtons to three significant figures. hints fg = 889 n submitmy answersgive up correct significant figures feedback: your answer 890 n was either rounded differently or used a different number of significant figures than required for this part. part c what is the gravitational force fg between the two objects described in part b if the distance between them is only 50 km ?
Physics
2 answers:
kobusy [5.1K]3 years ago
8 0
<span>b) The force with a distance of 150 km is 889 N c) The force with a distance of 50 km is 8000 N This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question. Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as F = (G M1 M2)/r^2 where F = Force G = gravitational constant M1 = Mass 1 M2 = Mass 2 r = distance between center of masses for the two masses. So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889. Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits. If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
rusak2 [61]3 years ago
7 0

(b). The gravitational force between the objects when they are 150 km apart is \boxed{889\,{\text{N}}}.

(c). The gravitational force between the objects when they are 50 km apart is \boxed{8000\,{\text{N}}} .

Further Explanation:

The gravitational force of attraction between the two bodies is given by the Newton’s Law of Gravitation. According to Newton’s law of Gravitation, the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The gravitational force is expressed mathematically as:

F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}

Here, G is the gravitational constant, {m_1} is the mass of first body, {m_2} is the mass of second body and r is the distance between two bodies.

For two bodies kept at a distance of 100 Km , the gravitational force of attraction is 2000 N.

Substitute 200 N for F and 100 Km for r in above equation.

\begin{aligned}2000=\frac{{G{m_1}{m_2}}}{{{{\left( {100\times {{10}^3}}\right)}^2}}}\hfill\\G{m_1}{m_2} = 2\times {10^{13}}\hfill\\\end{aligned}

<u>Part (b):</u>

Now, the force experienced by the bodies when they are 150 Km apart is:

F = \dfrac{{G{m_1}{m_2}}}{{\left( {150 \times {{10}^3}}\right)}}

Substitute 2 \times{10^8} for G{m_1}{m_2} in above equation.

\begin{aligned}F&=\frac{{2 \times {{10}^{13}}}}{{{{\left( {150 \times {{10}^3}}\right)}^2}}}\\&= 888.9\,{\text{N}}\\&\approx {\text{889}}\,{\text{N}}\\\end{aligned}

Thus, the gravitational force between the objects when they are 150 Km  apart is \boxed{889\,{\text{N}}}.

<u>Part (c): </u>

Now, the force experienced by the bodies when they are 50 km apart is:

F =\Dfrac{{G{m_1}{m_2}}}{{\left( {50 \times {{10}^3}}\right)}}

Substitute 2 \times {10^8} for G{m_1}{m_2} in above equation.

\begin{aligned}F &=\frac{{2 \times {{10}^{13}}}}{{{{\left({50 \times {{10}^3}}\right)}^2}}}\\&= 8000\,{\text{N}}\\\end{aligned}

Thus, the gravitational force between the objects when they are 50 Km  apart is  \boxed{8000\,{\text{N}}}.

Learn More:

1. Calculate the total force on the earth due to Venus, Jupiter, and Saturn brainly.com/question/2887352

2.Compare the surface area–to–volume ratios of Earth and Venus brainly.com/question/7227193

3.A rocket being thrust upward as the force of the fuel being burned brainly.com/question/11411375

Answer Details:

Grade: College

Subject: Physics

Chapter: Newton’s law of Gravitation

Keywords:  Gravitation, newton’s law, force of attraction, 2000N, 100km, 150 km, 50 km, gravitational force, two spherical objects, 889 N, 8000 N.

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Suppose an object moves along the y axis so that its location is yequals=​f(x)equals=xsquared2plus+x at time x​ (y is in​ meters
Volgvan

Answer:

A) The average velocity is 144 m/s.

B) The average velocity is (111 + h) m/s.

C) The instantaneous velocity is 111 m/s.

Explanation:

The position of the object is given by the following function:

y = f(x) = x² + x

A) The average velocity can be calculated as follows:

av = (f(xf) - f(x0)) / (xf - x0)

Where:

av = average velocity

f(xf) = the value of the function at x = x-final (xf)

f(x0) = the value of the function at x = x-initial (x0)

xf = x-final, final value of "x"

x0 = x-initial, initial value of "x"

Then the averge rate of change from x0 = 55 to xf =88 will be:

av = (f(88) - f(55))/(88 - 55)

f(x) = x² + x

Evaluating the function in x = 88 and x = 55:

f(88) = 88² + 88 = 7832 m

f(55) = 55² + 55 = 3080 m

Then:

av = 7832 m - 3080 m / (88 s - 55 s) =<u> 144 m/s</u>

B) av = (f(55 + h) - f(55)) / (55 + h - 55)

Evaluating the function in x = 55 +h:

f(55+h) = (55+h)² + (55+h)

f(55+h) = (55+h) · (55+h) + (55 + h)

f(55+h) = 55² + 55h +55h + h² + 55 +h

Evaluating the function in x = 55

f(55) = 55² + 55

Then f(xf) - f(x0):

f(55+h) - f(55) = 55² + 55h +55h + h² + 55 +h - 55² - 55

f(55+h) - f(55) = h(111 + h)

Then:

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C) The instantaneous velocity is obtained by derivating the function and evaluating the derivative at x = 55.

Then:

f´(x) = 2x + 1

f´(55) = 2 · 55 + 1 = <u>111 m/s</u>

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