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4 years ago

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A 38,500 kg sphere is located 2.55 m from a 15,400 kg sphere. What is the gravitational force, rounded to the nearest thousandth

of a Newton, between the two spheres?

1 answer:

kkurt [141]4 years ago

3 0

The gravitational force between the two spheres is 0.006 N

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

For the two spheres in this problem, we have

$m_1 = 38,500 kg$

$m_2 = 15,400 kg$

r = 2.55 m

Substittuting into the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(38,500)(15,400)}{2.55^2}=0.006 N$

Learn more about gravitational force:

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A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

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Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

2) The <em>potential energy</em> at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

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