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3 years ago

1) A hot, bright star would appear in which section of an H-R diagram ?

Physics

2 answers:

Rudiy273 years ago

4 0

Answer 1) Option C) Upper Left

Explanation : Hertzsprung-Russell (or H-R) diagrams are a graphical tool used to classify the stars according to their brightness (luminosity), spectral type and hotness (temperature) . In this diagram the stars which are found to be hot and bright are placed on the top left upper corner. Stars which are in the stable phase of hydrogen burning lie along the Main Sequence according to their mass.

Answer 2) Option B) for about 90% of their lifetimes

Explanation : It is found that the main sequence stars spend their 90% of lifetime fusing hydrogen into helium. These main sequence stars burns the hydrogen present in its core, until it reaches the end of its life cycle. At this point, it leaves the main sequence.

Hence, It spends 90% of its lifetime in the main sequence.

Zepler [3.9K]3 years ago

3 0

1. C. Upper Left
2. B. For about 90% of their lifetime

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Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0

DENIUS [597]

Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

We have given mass m = 1.7 kg

radius of the circle r = 0.48 m $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

Kinetic energy is given 14 J

Kinetic energy is equal to $KE=\frac{1}{2}mv^2$

So $\frac{1}{2}\times 1.7\times v^2=14$

$v^2=16.47$

v = 4.05 m/sec

Centripetal force is equal to $F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N$

So tension in the string will be equal to 58.33 N ( this will be the strength of the string )

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3 years ago

A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if

Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

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3 years ago

A football player runs from his own goal line to the opposing team's goal line returning to the fifty yard line all i 18.0s calc

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<span>assuming the pitch is 100yards long, the player runs 100yards to the other goal then a further 50 yards back to the 50-yard line. So he/she runs 150yards in 18s
150/18 = 8.33yards per second average speed.
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Vav = (Vinitial+Vfinal)/2
Vav = 4.16m/s</span>

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3 years ago

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If it requires 5.0 j of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re

ANEK [815]

15.277.. j. I did the problem using a proportion. an additional 3.7m to the current 1.8 cm=5.5cm.

Therefore, 5.0 j/1.8cm=x/5.5cm

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3 years ago

The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magni

masya89 [10]

Answer:

B/4

Explanation:

The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

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2 years ago