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Vsevolod [243]
3 years ago
10

Why is it important that a satellite in orbit be above the earth's atmosphere?

Physics
1 answer:
Montano1993 [528]3 years ago
6 0

Explanation:

A stone is thrown upward neglecting air resistance. ... Why is it important that a satellite remain above Earth's atmosphere? The air resistance will slow it down and burn it. What happens to a satellite close to Earth's surface if it is given a speed exceeding 11.2 km/s?

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1. Applied research observational evidence 2. Basic research the experimental factor that changes in response to a change in the
lesya [120]

Answer:

This question is about matching each definition with its correct term. Please find the term matched with their appropriate definition below.

Explanation:

1. Empirical evidence: An empirical evidence is an observational evidence i.e an evidence gathered by observation or use of senses.

2. Dependent variable: Dependent variable is an experimental factor that changes in response to a change in the independent variable. In other words, it is dependent on the independent variable.

3. Applied research: Applied research is a type of research oriented at solving a present problem or need. It encompasses the production of products that can be sold for profit.

4. Hypothesis: A hypothesis in an experiment is a proposed explanation for a scientific problem that itself can be tested by experimentation. A hypothesis aims at providing a testable explanation to an observed problem.

5. Control: A control is a quantity in an experiment that remains unchanged or constant. It is kept the same by the experimenter for all groups in the experiment in order not to influence the outcome.

6. Basic research: Basic research is the research that expands knowledge in a particular area. It is the kind of research that aims at filling a knowledge void or satiating curiosity.

7. Independent variable: The independent variable is the experimental factor that is changed or manipulated deliberately by the scientist.

8 0
3 years ago
Difference between discrete and continuous charge distribution?
Vlada [557]
Discrete systems are those systems in which are made up of finite component particles a which are non-homogeneously arranged such that no smooth variation exists. It is such that all constituent particles have properties which vary randomly. They are direct opposite to continuous systems, which are smooth arrangement of particles which cannot be individually taken into consideration.
Was this answer helpful
5 0
3 years ago
Read 2 more answers
Magnet A doesn't have its poles labeled, but Magnet B has a clearly labeled north and south pole. If the
dusya [7]

Answer:

D:   The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole

Explanation:

D:  The side of Magnet A that's attracted to Magnet B's south pole must be Magnet A's north pole because

1) opposite poles attract each other

2) similar poles repel each other

3)magnetic lines of force start at the north pole and end at the south pole

4 0
3 years ago
Read 2 more answers
Topic Gravitational force amd firld strength.. help me please
I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

5 0
3 years ago
Could dark matter be a higher dimensional object? Why?
REY [17]
Well dark matter is , in a layman terms just a concept used to identify missing mass in the universe. (extremely simplified and probably wrong but look that up). It could be but then if it is then we would still be able to percive the matter in some form. All we know from instruments is that they have mass and absorb light. We would need more info before any conclusions but it is 100% a possibility. But remember that when someone says "3-d" for example they are talking about a certain space.. technically matter exists on all planes (at least all the planes we know of in our local universe.) The laws of the cosmos are not yet proven to be constant and even if they are what of outside it?

When you take high school physics ask your teacher.
3 0
3 years ago
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