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3 years ago

Nuclear fusion occurs in stars. a. True b. False

2 answers:

8 0

Stars are powered by nuclear fusion in their cores, mostly converting hydrogen into helium. The production of new elements via nuclear reactions is called nucleosynthesis. A star's mass determines what other type of nucleosynthesis occurs in its core (or during explosive changes in its life cycle). READ THIS AND YOU WILL UNDERSTAND I THINK IS TRUE

earnstyle [38]3 years ago

8 0

Answer : The given statement is True.

Explanation -

Nuclear fusion is a thermonuclear reaction in which light stable nuclei combine with each other in order to form a heavy stable nucleus with the release of large amount of energy.

In case of stars, four hydrogen nuclei combine with each other to form a helium nucleus with a release of 26 Mev per cycle.

Thus, nuclear fusion occurs in stars is true statement.

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Two ice skaters stand in the middle of an ice rink. Drew has a mass of 75 kg, and Lily has a mass of 55 kg. Drew holds Lily, and

ss7ja [257]

PART a)

Before Drew throw Lily in forwards direction they both stays at rest

So initial speed of both of them is zero

So here we can say that initial momentum of both of them is zero

So total momentum of the system initially = ZERO

PART b)

Since there is no external force on the system of two

so there will be no change in the momentum of this system and it will remain same as initial momentum

So final momentum of both of them will be ZERO

PART c)

As we know that momentum of both will be zero always

so we have

$P_1 + P_2 = 0$

$75(v) + 55(2) = 0$

$v = 1.47 m/s$ in opposite direction

7 0

3 years ago

Read 2 more answers

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

What is required for both the light-dependent and light-independent reactions to proceed?

djyliett [7]

ATP is required for both light-dependent and light-independent reactions.
ATP stands for adenosine triphosphate.
Hope this helps ;)

3 0

3 years ago

Body composition refers to:

NISA [10]

Bone, fat and muscle

3 0

2 years ago

A 47.2 kg girl is standing on a 177 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

Softa [21]

Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

Mass of girl; m_g = 47.2 kg

Mass of plank; m_p = 177 kg

Let the velocity of girl to ice be v_g,i

Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

v_g,i + v_p,i = 1.53

From conservation of momentum;

m_g × v_g,i = m_p × v_p,i

Thus;

47.2(v_g,i) = 177(v_p,i)

Dividing both sides by 47.2 gives;

v_g,i = 3.75(v_p,i)

v_pi = (v_g,i)/3.75

Thus, from v_g,i + v_p,i = 1.53, we have;

v_g,i + ((v_g,i)/3.75) = 1.53

v_g,i(1 + 1/3.75) = 1.53

1.267v_g,i = 1.53

v_g,i = 1.53/1.267

v_g,i = 1.208 m/s

5 0

3 years ago