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emmainna [20.7K]
3 years ago
9

How old is a bone if it still has 50% of its carbon-14 content?

Physics
1 answer:
Degger [83]3 years ago
6 0

Answer:

5730 years

Explanation:

The half life of carbon-14 is 5730 years.  If 50% of the carbon-14 remains, then exactly 1 half life has passed.

The half-life equation is:

A = A₀ (½)^(t / T)

where A is the remaining amount,

A₀ is the initial amount,

t is time,

and T is the half life.

In this case, A = ½ A₀ and T = 5730.

½ A₀ = A₀ (½)^(t / 5730)

½ = (½)^(t / 5730)

1 = t / 5730

t = 5730

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A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a
Kaylis [27]

Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

W = 0.3853 J

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3 years ago
Where might you look on the internet to find good scientific information about illness
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3 years ago
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What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change
sammy [17]

Answer:

0.1 m

Explanation:

The closest distance the electrodes used in an NCV test in oerder to measure

the voltage change as a response to the stimulus is 0.1 m.

This is because the shortest observable time period is not less than the action-potential time response of 1 mili second the length traveled by the sensation during this time is 1 m sec x 100 m / s =0.1 m, which is the shortest distance the electrodes could be positioned on the nerve.

6 0
3 years ago
A 91.0-kg hockey player is skating on ice at 5.50 m/s. another hockey player of equal mass, moving at 8.1 m/s in the
never [62]

The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.

(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

1237.6 kg m/s² =(2 ×91.0-kg )V

V=6.8 m/sec.

Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

8 0
2 years ago
Fill in the blank. Consider the inverse square law: When light leaves a light bulb, it spreads out over more and more space as i
aliya0001 [1]

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r²  where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

"16 times " is the answer .

8 0
3 years ago
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