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3 years ago

Why is iron significant to understanding how a supernova occurs?

Physics

1 answer:

Vilka [71]3 years ago

6 0

<h3><u>Answer;</u></h3>

Iron cannot release energy by fusion.

<h3><u>Explanation;</u></h3>

Supernovae are some of the most magnificent events in the cosmos. When a massive star dies, it can result in a spectacular explosion many times that of man's total nuclear arsenal.
When a star has used up all the lighter elements and has just iron left, it has no more nuclear "fuel". That causes the star to contract then explode very violently as a supernova.

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An air bubble is at a depth of 3 m below

lisabon 2012 [21]

Answer:

incorrect its 987 for exact

4 0

2 years ago

A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude

miskamm [114]

Answer:

Explanation:

Given

Weight of person $W=652\ N$

At highest point Magnitude of the normal force $F=585\ N$

net force at highest point

$F_{net}=W+F_c$

where $F_c=$ centripetal force

$F_c=\dfrac{mv^2}{r}$

$F_{net}=F_{n}=$ Normal Force

$585=652+F_c$

$F_c=-67\ N$

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

$F_n=F_c+W$

$F_n=652+67$

$F_n=719\ N$

8 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$