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2 years ago

11

An object weighs 200 newtons at a distance of 100 kilometers above the center of a small uniform planet. how much will the objec

t weigh 200 kilometers above the planet's center?

1 answer:

disa [49]2 years ago

7 0

Since the law of gravitation is an inverse square law if you quadruple the radius the f will drop by a factor of 16 SO the object would weigh 200/16 = 12.5N

In other words, as the distance, or radius, quadruples the weight becomes 1/16 of the original weight. Just plug in 4 for r and when you square it you get 16. The numerator is 1 so that is how the weight becomes 1/16.

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Answer:

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Explanation:

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2 years ago

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Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was appro

Masteriza [31]

Answer:

61578948 m/s

Explanation:

λ $_{actual}$ = λ $_{observed}$ $\frac{c+v_{o}}{c}$

687 = 570 $(\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )$

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So Slick Willy was travelling at a speed of 61578948 m/s to observe this.

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2 years ago

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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2 years ago

For which of the following problems would a scientist most likely use carbon-14?

spayn [35]

Answer:

To calculate the age of a piece of bone

Explanation:

Carbon 14 is an isotope of carbon that is unstable and decays into Nitrogen 14 by emitting an electron. The decay rate of radioactive material is normally expressed in terms of its "half-life" (the time required by half the radioactive nuclei of a sample to undergo radioactive decay). The nice thing about carbon 14 is that its "half-life" is about 5730 years, which gives a nice reference to measure the age of fossils that are some thousand years old.

Carbon 14 dating is used to determine the age of objects that have been living organisms long ago. They measure how much carbon 14 is left in the object after years of decaying without having exchange with the ambient via respiration, ingestion, absorption, etc. and therefore having renewed the normal amount of carbon 14 that is in the ambient.

A rock is not a living organism, so its age cannot be determined by carbon 14 dating.

3 0

2 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun

AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

$i_1+i_3=i_2$ (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

$i_3=i_2-i_1=0.75A-0.40A=0.35A$

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

$0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}$

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

3 0

2 years ago