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baherus [9]
2 years ago
14

Although centuries ago, astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we kn

ow that it is:
Physics
1 answer:
Nutka1998 [239]2 years ago
5 0

Astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is as a binary star system.

<h3>What is the binary star system about?</h3>

A binary star system is known to be one where one star is known to be called a white dwarf and there is a mass that is said to be transferred to it

A binary star is known to be a kind of a system that is composed of two stars that are known to be gravitationally held together to and in orbit near each other.

Note that Binary stars in the night sky are ones that are often seen as a single object and thus Astronomers thought that a nova was a new star, appearing for the first time in the heavens, today we know that it is as a binary star system.

Learn more about astronomers  from

brainly.com/question/1141458

#SPJ1

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When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere
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What do the best conductors have in common?
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Answer:

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When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea
crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = \frac{(172\ \mu C)}{V}

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = \frac{(172+220)\ \mu C)}{V}

or

C' = \frac{(392)\ \mu C)}{V}

now the dielectric constant (k) is given as:

k=\frac{C'}{C}

substituting the values, we get

k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}

or

k = 2.279

6 0
3 years ago
How much work is requried to uniformly accelerate a merry-go-round?
tiny-mole [99]
Do not worry if you don't recognize both parts of the problem at this point. If you recognize the dynamics problem,<span> On the other hand, if you recognize this as a kinematics problem you will quickly see that you need to find angular acceleration before you can begin and so will need to do that pre-step first.</span>
5 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
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