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Fudgin [204]
3 years ago
11

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

where G is the gravitational constant and r is the distance between the bodies. (a) Find dF/dr. dF dr = − 2GmM r3​ What is the meaning of dF/dr? dF/dr represents the rate of change of the mass with respect to the distance between the bodies. dF/dr represents the amount of force per distance. dF/dr represents the rate of change of the mass with respect to the force. dF/dr represents the rate of change of the distance between the bodies with respect to the force. dF/dr represents the rate of change of the force with respect to the distance between the bodies. What does the minus sign indicate? The minus sign indicates that as the distance between the bodies increases, the magnitude of the force decreases. The minus sign indicates that the bodies are being forced in the negative direction. The minus sign indicates that the force between the bodies is decreasing. The minus sign indicates that as the distance between the bodies increases, the magnitude of the force increases. The minus sign indicates that as the distance between the bodies decreases, the magnitude of the force remains constant. (b) Suppose it is known that the earth attracts an object with a force that decreases at the rate of 4 N/km when r = 20,000 km. How fast does this force change when r = 10,000 km? N/km
Physics
1 answer:
Vladimir [108]3 years ago
8 0
<h2>Answers:</h2>

<h2><u>Answer 1 (a): </u></h2><h2> </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 <u>are constants:</u>

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

Or

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

<h2><u>Answer 2 (a):</u>  dF/dr represents the rate of change of the force with respect to the distance between the bodies.  </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.

<u>As the distance increases, the Force decreases.</u>

<h2><u>Answer 3 (a):</u>  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><u>Answer 4 (b):</u>  X=-32N/km</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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At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
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Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
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