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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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if an object is moving at a constant speed in a changing direction, What is the acceleration of the object (zero or not zero)​
Andreas93 [3]
  • We know, acceleration is the change of velocity by time.
  • Velocity is the speed of an object which also indicates the direction.
  • Hence, acceleration is both dependant upon the speed as well as the direction.
  • So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
  • An example is that of uniform circular motion.

Answer:

if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.

4 0
3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
An ice skater accelerates backward for 5.0 second to final speed of 12m/s. If the acceleration backward was at a a rate of 1.5m/
Lisa [10]

Answer:

Initial velocity, U = 4.5m/s

Explanation:

Given the following data;

Final velocity, v = 12m/s

Time, t = 5 seconds

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To find the initial velocity, we would use the first equation of motion.

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the equation, we have;

12 = U + 1.5*5

12 = U + 7.5

U = 12 - 7.5

Initial velocity, U = 4.5m/s

3 0
3 years ago
"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

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Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

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so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

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N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

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