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2 years ago

The layers in a high-mass star occur in order of:________

Physics

1 answer:

zavuch27 [327]2 years ago

6 0

A high-mass star has layers that are arranged in order of b. Temperature for fusion.

<h3>What is the temperature of fusion?</h3>

The temperature at which a substance transforms from a solid to a liquid is known as its melting point. The solid and liquid phases are in equilibrium at the melting point. Pressure affects a substance's melting point, which is typically reported at a standard pressure such 1 atmosphere or 100 kPa. There are two main categories of fusion reactions: those that maintain the ratio of protons to neutrons and those that convert protons to neutrons. The solid-liquid and liquid-to-gas transitions are connected with the typical heats of fusion and vaporization, respectively, for each substance.

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Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, $B=2\ mT=2.2\times 10^{-3}\ T$

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :

$B=\dfrac{\mu_oNI}{2\pi r}$

$r=\dfrac{\mu_oNI}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}$

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2 years ago

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A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

$y=y_{0}+uSin\theta t-1/2 gt^{2}$

$y=2+20Sin5 t-4.9t^{2}$

As it hits the ground in time t, so put y = 0

$0=2+1.74 t-4.9t^{2}$

$4.9t^{2}-1.74t-2=0$

$t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}$

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

$y'=y_{0}+uSin\theta t'-1/2 gt'^{2}$

$y'=2+20Sin5 t-4.9\times0.35^{2}$

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

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