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fomenos
3 years ago
13

Testing shows that a sample of wood from an artifact contains 25% of the

Physics
2 answers:
daser333 [38]3 years ago
5 0

Answer:

The artifact is 11,460 years old.

<u>Explanation</u>:

Given that,

The half life of the carbon-14 is 5730  years and we are left with 255 of the sample of wood from an arti-fact.

So it takes 5730 years for the sample to reduce into half

Initially there will be 100% of the sample so

after first 5730 years, the sample reduces into 50% percent

Now the left 50% sample will take another 5730 years to decay into half of its amount.

after next 5730 years  the sample reduces into 25% percent

So totally after 2 half-life the sample reduces into 25%

That is (5730 +5730) years = 11460 years

Umnica [9.8K]3 years ago
3 0

Answer:5730 years

Explanation:

A P E X

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A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr
quester [9]

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = \frac{n}{2L} x \sqrt{\frac{T}{μ}}

f = \frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

8 0
3 years ago
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
2560 meters in 60 seconds, what is his average speed (velocity)
jok3333 [9.3K]
42.6 is the answer I believe because you would do 2,560 divided by 60 if I'm correct.
5 0
3 years ago
Find the intensity of a 55 dB sound given I 0=10^-12W/m^2
MakcuM [25]

Answer:

3.16 × 10^{-7} W/m^{2}

Explanation:

β(dB)=10 × log_{10}(\frac{I}{I_{0} })

I_{0}=10^{-12} W/m^{2}

β=55 dB

Therefore plugging into the equation the values,

55=10 log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

5.5= log_{10}(\frac{I}{ [tex]10^{-12}})[/tex]

10^{5.5}= \frac{I}{10^{-12} }

316227.76×10^{-12}= I

I= 3.16 × 10^{-7} W/m^{2}

5 0
3 years ago
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