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Zinaida [17]
2 years ago
12

carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr

oximately how many years ago was this shell part of a living oranism?
Physics
1 answer:
denis23 [38]2 years ago
3 0

The half-life equation m=m_{0} (\frac{1}{2})^n in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>

<em>n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}</em>

<em>\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2</em>

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

<em>2 * 5700 =  </em>11400 yr

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Can a small child play with fat child on the seesaw?Explain how?
RoseWind [281]
Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play
8 0
2 years ago
Read 2 more answers
While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the
olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0
2 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
2 years ago
Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,
VARVARA [1.3K]

Answer: yes.

Explanation: The light that will be incidented on that metal is visible light.

It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer. 

6 0
2 years ago
A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions
erastova [34]
•wind
•snow
•high tide/low tide
•thunder/lightning storms
5 0
2 years ago
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