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Sauron [17]
3 years ago
6

Why is the definition for second based on the Cesium 133 atom?

Engineering
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer:

  The rate of vibrations of the atoms is consistent

Explanation:

A second is defined as 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

The definition applies when the atom is in isolation at a temperature of absolute zero. Adjustments must be made for the conditions that can be maintained in practice.

Such radiation is determined by the laws of quantum mechanics, so is extremely consistent.

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Thermosets burn upon heating. a)-True b)- false?
hram777 [196]

Answer:

true

Explanation:

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3 years ago
Calculate the magnitude of the velocity and the θ angular direction of the block and the bullet together when the 50 g bullet mo
almond37 [142]

Answer:

Magnitude of the velocity = 16.82 m/s

Angular direction, θ = 52.41°

Explanation:

As given ,

mass of bullet, m₁= 50g = 0.05 kg

speed of bullet , u₁ = 600 m/s

mass of the block , m₂ = 4 kg

speed of the block before collision , u₂ = 12 m/s

direction , θ = 30°

Now,

Assume that the combined velocity of bullet and block after collision = v

and the direction = θ

Now, from the conservation of momentum in x - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ

where v = final velocity after collision

u₁ = initial velocity of bullet before collision = 0

m₁ = mass of the bullet before collision = 0.05 kg

u₂  = velocity of block before collision = 12 cos(30° )

m₂ = mass of block before collision

m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ

⇒ 0 + 4(6√3) = 4.05 vₓ

⇒24√3 = 4.05 vₓ

⇒vₓ = 10.26 m/s

Now, from the conservation of momentum in y - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) v_{y}

where v = final velocity after collision

u₁= initial velocity of bullet before collision = 600

m₁ = mass of the bullet before collision = 0.05 kg

u₂  = velocity of block before collision = 12 sin(30° )

m₂= mass of block before collision

m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 ) v_{y}

⇒ 30 + 4(6) = 4.05 v_{y}

⇒30 +24 = 4.05 v_{y}

⇒54 = 4.05 v_{y}

⇒v_{y} = 13.33 m/s

Now, the magnitude of the velocity = √vₓ² + v_{y}² = √(10.26)² + (13.33)²

                                                           = √105.26 + 177.68

                                                           = √282.95 = 16.82

The angular direction, θ =  tan^{-1}(\frac{v_{y} }{v_{x} }) =  tan^{-1}(\frac{13.33}{10.26}) = tan^{-1}(1.299) = 52.41°

8 0
3 years ago
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