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3 years ago

If you had started with a larger mass, how would the half-life change?

Physics

1 answer:

iogann1982 [59]3 years ago

3 0

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

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In nuclear fission the nucleus is divided into two or more fragments, releasing Question 8 options: protons and energy. neutrons

Dmitrij [34]

Answer:

the answer is neutrons and energy

8 0

3 years ago

A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, How far will it have traveled horizontally?

Maurinko [17]

Distance traveled by the ball is given by

$distance = speed \times time$

here we know that

speed = 20 m/s

times = 0.25 s

now we have

$distance = 20 \times 0.25$

$distance = 5 m$

so ball will travel 5 m distance in the given interval of time

6 0

3 years ago

A simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top

Trava [24]

Answer:

He will complete the race in total time of T = 10 s

Explanation:

Total distance moved by the sprinter in 2.14 s is given as

$s = \frac{(v_{in} + v_{f})}{2} time$

$s = \frac{(0 + 11.2)}{2} (2.14)$

$s = 11.98 m$

now the distance remaining to move

$d = 100 - 11.98 = 88 m$

now he will move with uniform maximum speed for the remaining distance

so we will have

$time = \frac{d}{v}$

$time = \frac{88}{11.2} = 7.86 s$

so the total time to complete the race is given as

$T = 7.86 + 2.14 = 10 s$

6 0

3 years ago

Could someone please help me

BaLLatris [955]

Answer:

the size of the shadow will be smaller due to smaller hands

3 0

3 years ago

when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2

Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂ = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0

3 years ago