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3 years ago

A planet has two

1 answer:

7 0

Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

Moon #1: (1.262 days)² / (2.346 x 10^4 km)³

Moon #2: (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

= 0.1017 day²

Orbital period = 0.319 Earth day = about 7.6 hours.

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Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)

Marat540 [252]

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

$V=\frac{Kq}{r}$

Then, the net potential is:

$V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}$

$K=\frac{1}{4\pi \epsilon _{o} }$

$V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0$

8 0

4 years ago

Which two phrases describe critical thinking skills used in the pursuit of science?

SSSSS [86.1K]

Answer:

A. Expanding personal knowledge by reading articles from scientific

journals

B. Analyzing the different parts of a physical phenomenon to see

how they fit together

C. Predicting the impact that answering an important scientific

question would have on people

D. Developing a question that can be answered through testing or

observation

E. Developing logical arguments for providing incentives for

scientific research

6 0

3 years ago

At t = 0, one toy car is set rolling on a straight track with initial position 13.5 cm, initial velocity -4.2 cm/s, and constant

Blizzard [7]

(a) We must first look at the formulas of the velocities of each toy car. v1 =
-4.2 + 2.60t. v2 = 5.20. When the two cars have equal speed, then
v1 = v2
-4.2 + 2.60t = 5.20
2.60t = 9.40
t = 3.62 s

(b) Their speed would then be 5.20 m/s. The toy car does not change speed since it doest not have any acceleration.
(c) The two cars will pass each other when their positions are equal.
x1 = 13.5 - 4.2t + 0.5*2.60t^2
x2 = 8.5 + 5.20t
x1 = x2
13.5 - 4.2t + 1.30t^2 = 8.5 + 5.20t
1.30t^2 - 9.40t + 5.0 = 0
t = 6.65s or t = 0.58 s

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I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

3 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago

What is diffrence between damping and undamping?

ICE Princess25 [194]

Answer:

Oscillation whose amplitude reduce with time are called damped oscillation. This happen because of the friction. In oscillation if its amplitude doesn't change with time then they are called Undamped oscillation

7 0

3 years ago

Read 2 more answers