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amid [387]
2 years ago
7

What physical properties dose plutonium

Physics
1 answer:
Minchanka [31]2 years ago
6 0

Answer:

Explanation:

Plutonium-239 is one of the two fissile materials used for the production of nuclear weapons and in some nuclear reactors as a source of energy. The other fissile material is uranium-235. Plutonium-239 is virtually nonexistent in nature. It is made by bombarding uranium-238 with neutrons in a nuclear reactor. Uranium-238 is present in quantity in most reactor fuel; hence plutonium-239 is continuously made in these reactors. Since plutonium-239 can itself be split by neutrons to release energy, plutonium-239 provides a portion of the energy generation in a nuclear reactor.

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Urgent!
Masja [62]

mass of iron block given as

m_1 = 1.90 kg

density of iron block is

\rho = 7860 kg/m^3

now the volume of the iron piece is given as

V = \frac{m}{\rho}

V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

F_b = \rho_L V g

here we know that

\rho_L = density of liquid = 916 kg/m^3

F_b = 916* 2.42 * 10^{-4} * 9.8

F_b = 2.17 N

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

F_s + F_b = mg

F_s + 2.17 = 1.90* 9.8

F_s = 16.45 N

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

F_n = F_g + F_b

F_n = (1 + 2.50)*9.8 + 2.17

F_n = 34.3 + 2.17 = 36.47 N

So the other scale will read 36.47 N

3 0
3 years ago
When passing from a low index of refraction medium to a high index of refraction medium which way does the refracted ray bend: t
Vaselesa [24]

Answer:

toward the normal

Explanation:

Light travels at different speed in different mediums.

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.  

When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.

6 0
3 years ago
The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is
Fudgin [204]

Answer:

24cm/s

Explanation:

A=L*w

A'=L'*w'

L=13

w=5

L'=4

w'=6

A=?

A'=?

A=L*w

A=13*5

A=65

A'=L'*w'

A'=4*6

A'=24

*the given lengths are just to throw you off*

3 0
3 years ago
Astronomers use a special kind of instrument called a spectrometer to analyze stars. Spectrometers can determine what stars are
marshall27 [118]
The approval of the stars
4 0
2 years ago
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
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