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1 year ago

10. Complete each of the following radioactive decay equations.

Physics

1 answer:

alukav5142 [94]1 year ago

6 0

1. Alpha

Number next to Th is 231

(because 231 + 4 = 235)

The other number below the 4 is 2 - helium

(because 90 + 2 = 92)

2. Beta

Missing numbers are zero and one. Same as the one below it.

3. Beta

11 is the number above the C

(because 11 + 0 = 11)

5 is the number next to the B

(because 5 + 1 = 6)

Basically, the missing numbers must balance on both side of the equation; creating the elements you've started of with, meaning if you where to add the right side of the equation up - you should end up with the same protons and mass numbers you began with.

Hope this helps!

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You pull with a force of 77 N on a piece of luggage of mass 23 kg, but it does

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Answer:

The force of static friction acting on the luggage is, Fₓ = 180.32 N

Explanation:

Given data,

The mass of the luggage, m = 23 kg

You pulled the luggage with a force of, F = 77 N

The coefficient of static friction of luggage and floor, μₓ = 0.8

The formula for static frictional force is,

Fₓ = μₓ · η

Where,

η - normal force acting on the luggage 'mg'

Substituting the values in the above equation,

Fₓ = 0.8 x 23 x 9.8

= 180.32 N

Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N

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Instructions:Select all the correct answers.

melomori [17]

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Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls

Alina [70]

Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 = 8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

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