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3 years ago

An earth satellite in an elliptical orbit travels slowest when it is

1 answer:

8 0

The satellite travels slowest when it is at the maximum distance from the Earth.

We can verify this in two ways:

1) By using Kepler's second law: "A line segment joining a a satellite with the Earth covers equal areas during equal intervals of time". This means that the larger is the distance of the satellite from Earth, the slower it goes.

2) by looking at the forces acting on the satellite. There is only one force acting on it: the gravitational attraction exerted by Earth, and this force is the centripetal force that keeps the satellite in circular (elliptical, actually) motion. So we can write:
$G \frac{Mm}{r^2}=m \frac{v^2}{r}$
where on the left we wrote the formula of the gravitational force, while on the right the centripetal force. G is the gravitational constant, M the Earth's mass, m the satellite's mass, v its velocity and r the distance of the satellite from the center of Earth.
Simplifying, we get
$v= \sqrt{ \frac{GM}{r} }$
which is the speed of the satellite when it is at a distance r from Earth: the larger r, the smaller the speed v.

You might be interested in

Define Archemedics principle?

azamat

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.

6 0

3 years ago

A machine is designed to fill jars with 16 ounces of coffee. A consumer suspects that the machine is not filling the jars comple

babunello [35]

Answer:

a) Null hypothesis: $\mu \geq 16$

Alternative hypothesis : $\mu$

b) $t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77$

$p_v =P(t_{(7)}$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

c) We can conclude that the mean is significantly less than 16 ounces at 10% of significance. so then the consumer suspect is correct.

Explanation:

Data given and notation

$\bar X=15.6$ represent the mean for the sample

$s=0.3$ represent the sample standard deviation

$n=8$ sample size

$\mu_o =16$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

Is a one tailed left test.

What are H0 and Ha for this study?

Null hypothesis: $\mu \geq 16$

Alternative hypothesis : $\mu$

The degrees of freedom on this case are:

$df=n-1=8-1=7$

Compute the test statistic

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{15.6-16}{\frac{0.3}{\sqrt{8}}}=-3.77$

Give the appropriate conclusion for the test

Since is a one side left tailed test the p value would be:

$p_v =P(t_{(7)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can conclude that the mean is significantly less than 16 ounces at 10% of significance. so then the consumer suspect is correct.

3 0

3 years ago

“Anchor” points on a temperature scale are known as fiducial points.<br><br> a)True<br> b)False

Svetlanka [38]

Yes. It r<span>efers to any of the temperatures assigned to a number of reproducible equilibrium states on the International Practical Temperature Scale</span><span>

In short, Your Answer would be "True"

Hope this helps!</span>

8 0

3 years ago

Read 2 more answers

Please help me! Uniform acceleration problem sheet:

Brrunno [24]

From the calculation, the value of the acceleration is 5.8 m/s^2.

<h3>What is uniform acceleration?</h3>

The term uniform acceleration refers to a situation in which the velocity increases by equal amounts in equal time intervals.

Given the fact that the car started from rest and reached a velocity of 780.34 mph or 348.84 m/s in 1 minute of 60 seconds;

v = u + at

a = v/t

a = 348.84 m/s/ 60 seconds

a = 5.8 m/s^2

Learn more about acceleration:brainly.com/question/12550364?

#SPJ1

7 0

2 years ago

Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n

damaskus [11]

Answer:

38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
Distance covered by car A = speed x time(t) = 60 x t = 60t
Distance covered by car B = speed x time(t) = 75 x t = 75t
60t + 75t = 340 miles
135t = 340
t = 2.51 hours
Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0

3 years ago