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3 years ago

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Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?

1 answer:

IRISSAK [1]3 years ago

4 0

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

3 years ago

At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de

e-lub [12.9K]

Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.

Answer:

2.99 m/s

Explanation:

Stopping distance, s = 3 ft = 0.914 m

final velocity, v = 0

a = g/2 = 4.9 m/s²

Use third equation of motion:

$v^2-u^2 = 2as$

substitute the values to find the speed of train:

$0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s$

6 0

3 years ago

Stress distributed over an area is best described as: a) External force b) Axial force c) Radial force d) Internal resistive for

Anit [1.1K]

Answer:

Option D is the correct answer.

Explanation:

Stress is the force per unit area that tend to change the shape of body.

Stress is defined as internal resistive force per unit area.

$\texttt{Stress}=\frac{\texttt{Internal resistive force}}{\texttt{Area}}$

$\sigma =\frac{F}{A}$

So, so stress distributed over an area is best described as internal resistive force.

Option D is the correct answer.

8 0

3 years ago

PLEASE HURRYY!!!!The diagram shows two balls released from a device at the same time. The ball on the left falls freely from res

LenaWriter [7]

Answer:

i'm pretty sure its B but i may be wrong if you dont wanna take the chance wait for someone

Explanation:

4 0

3 years ago

Read 2 more answers

A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall

kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

$F = q(\vec v \times \vec B)$

so here we have

$F = qvB sin90$

since force is perpendicular to the velocity so here it must be centripetal force

here we have

$\frac{mv^2}{R} = qvB$

so we have

$R = \frac{mv}{qB}$

$R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}$

$R = 4.06 cm$

so here it will move in circle with radius 4.06 cm

8 0

3 years ago