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2 years ago

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A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam

ple remain after 17,100 years

2 answers:

ahrayia [7]2 years ago

8 0

Answer:

1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

(1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

agasfer [191]2 years ago

6 0

Answer:

1/8

Explanation:

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

Infer whether a system can have kinetic energy and potential energy at the same time

Rasek [7]

Yes, an object can have both of these at the same time. Potential energy is energy that is stored in an object. Kinetic energy is the energy that is associated with motion. So what you have to have is an object that is in motion but still has more energy that it has yet to convert into kinetic energy.

4 0

2 years ago

(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th

Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:

$V_{f}^{2} = V_{0}^{2} - 2gh$

<u>Where:</u>

$V_{f}$ : is the final speed = 14 m/s

$V_{0}$ : is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

$V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s$

b) The maximum height is:

$V_{f}^{2} = V_{0}^{2} - 2gh$

$h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m$

c) The time can be found using the following equation:

$V_{f} = V_{0} - gt$

$t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s$

d) The flight time is given by:

$t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s$

I hope it helps you!

3 0

3 years ago

How does the movement of particles of matter change when matter goes from solid to liquid? (1 point)

Y_Kistochka [10]

Answer: c

Explanation: because if it increases then it gets hotter

if it decreases then its getting colder.

3 0

2 years ago

The spring of a toy car is wound by pushing the car

Zepler [3.9K]

The elastic potential energy stored in the car's spring during the process is 3.75 J

<h3>Determination of the spring constant</h3>

From the question given above, the following data were obtained:

Force (F) = 15 N

Extention (e) = 0.5 m

Spring constant (K) =?

K = F/e

K = 15 / 0.5

K = 30 N/m

<h3>Determination of the potential energy</h3>

Spring constant (K) = 30 N/m

Extention (e) = 0.5 m

Potential energy (PE) =?

PE = ½Ke²

PE = ½ × 30 × 0.5²

PE = 15 × 0.25

PE = 3.75 J

Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J

Learn more about energy stored in spring:

brainly.com/question/4280346

7 0

2 years ago

Read 2 more answers