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3 years ago

The uranium in nuclear reactors comes from supernova explosions.

Physics

1 answer:

Ludmilka [50]3 years ago

4 0

Answer: this statement makes sense.

Explanation:

Supernova explosion is a bright powerful explosion of the start, it takes place at the end of the cycle of a star. This explosion is caused by having too much matter in it's core, As the star runs out of nuclear fuel, some of its mass flows into its core. Eventually, the core is so heavy that it cannot withstand its own gravitational force. The core collapses, which results in a massive explosion of a supernova. During this explosion Elements heavier than iron are formed. An uranium atom is heavier than an iron atom, so uranium falls into this category of elements that comes from supernova explosion.

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A gyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 rev/min.

SIZIF [17.4K]

Answer:

Part a)

$a_t = 0.423 m/s^2$

Part b)

$a_c = 2113 m/s^2$

Part c)

$d = 80 m$

Explanation:

Part a)

as we know that angular acceleration of the wheel is given as

$\alpha = 13.2 rad/s^2$

now the radius of the wheel is given as

R = 3.21 cm

so the tangential acceleration is given as

$a_t = R\alpha$

$a_t = (0.0321)(13.2)$

$a_t = 0.423 m/s^2$

Part b)

frequency of the wheel at maximum speed is given as

$f = 2450 rev/min$

$f = \frac{2450}{60} = 40.8 rev/s$

now we know that

$\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s$

now radial acceleration is given as

$a_c = \omega^2 r$

$a_c = (256.56)^2(0.0321) = 2113 m/s^2$

Part c)

total angular displacement of the point on rim is given as

$\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

here we know that

$\omega = \omega_0 + \alpha t$

$256.56 = 0 + 13.2 t$

$t = 19.4 s$

now angular displacement will be

$\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2$

$\Delta \theta = 2493.3 rad$

now the distance moved by the point on the rim is given as

$d = R\theta$

$d = (0.0321)(2493.3)$

$d = 80 m$

7 0

3 years ago

A ray of monochromatic light ( f = 5.09 × 10^14 Hz) passes from air into Lucite at an angle

harina [27]

Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:

$\frac{sen\O_{2}}{sen\O_{1}} = \frac{n_{2}}{n_{1}} \\ sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}}$

Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:

$sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}} \\ sen\O_{2}= \frac{1.5* \frac{1}{2} }{1} \\ sen\O_{2}=0.75$

Using the arcsin properties, we get:

$sen\O_{2}=0.75 \\ arcsin(0.75)=\O_{2} \\ \boxed {\O_{2}=48.59^o}$

Obs: Approximate results, and the drawing is attached

If you notice any mistake in my english, let me know, because i am not native.

6 0

3 years ago

Energy waves traveling at the speed of light are called _____.

koban [17]

RADIATION

There are three types of heat transfer or heat propagation; conduction, convection and radiation. Heat transfer is the process by which heat projects externally however, depending on the temperature and pressure. Also called the movement of heat from a low temperatured state which increases as heat progresses.
Conduction is the heat transfer by contact, immediate contact.
Convection is the transfer of heat through air and water.
<span>Radiation is the transfer of heat regardless of the presence of atoms or particles.<span>
</span></span>

8 0

3 years ago

Read 2 more answers

Which of the following best represents a chemical reaction?

Sonja [21]

Answer:

The answer to your question should be D.

Explanation:

reactants are on the laft side of arrow and products are on right side of arrow

7 0

3 years ago

The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s$

Work done is equal to the change in kinetic energy. It can be given as follows :

$W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J$

So, the required work done is 32.99 J.

3 0

3 years ago