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alexgriva [62]
2 years ago
13

What are the similarities and differences between novae and supernovae?

Physics
1 answer:
DochEvi [55]2 years ago
3 0

Answer:

Explanation:

A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.

A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.

From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae

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When an object is placed 110 cm from a diverging thin lens, its image is found to be 55 cm from the lens. The lens is removed, a
Kazeer [188]

Explanation:

Given that,

Object distance u= -110 cm

Image distance v= 55 cm

We need to calculate the focal length for diverging lens

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-f}=\dfrac{1}{55}-\dfrac{1}{-110}

\dfrac{1}{f}=-\dfrac{3}{110}

f=-36.6\ cm

The focal length of the diverging lens is 36.6 cm.

Now given a thin lens with same magnitude of focal length 36.6 cm is replaced.    

Here, The object distance is again the same.

We need to calculate the image distance for converging lens

Using formula of lens

\dfrac{1}{36.6}=\dfrac{1}{v}-\dfrac{1}{-110}

Here, focal length is positive for converging lens

\dfrac{1}{v}=\dfrac{1}{36.6}-\dfrac{1}{110}

\dfrac{1}{v}=\dfrac{367}{20130}

v=54.85\ cm

The distance of the image is 54.85 cm from converging lens.

Hence, This is the required solution.

5 0
3 years ago
Calculate the time period of simple pendulum whose length is 98.2cm
alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \  \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

#SPJ1

5 0
1 year ago
A basketball player can jump 1.6 m off the hardwood floor. With what upward velocity did he leave the floor?
irinina [24]
First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2

= 31.36

Sq root = 5.6
3 0
3 years ago
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mrs_skeptik [129]

Answer:

25 to the right

Explanation:

there you go friend your awsome

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2 years ago
[BWS.02]If the same experiment is repeated in different parts of the world by different scientists,
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the results will be the same.it may be

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