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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 long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

a)

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

3 years ago

The role of learning in motivation is most obvious from the influence of<br><br>

photoshop1234 [79]

Answer:

Your answer should be Incentives

Explanation:

7 0

2 years ago

A 98 N ball is suspended from a cable so that it hangs 3.5 m above the earth. Find the mass of the ball and the

expeople1 [14]

Answer:

Yes

Explanation:

4 0

2 years ago

a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road

Svetradugi [14.3K]

Answer:

t = 4.1 seconds

Explanation:

It is given that,

Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.

Speed of man is 2.01 m/s

We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

t is time taken

$t=\dfrac{d}{v}\\\\t=\dfrac{8.25}{2.01}\\\\t=4.1\ s$

So, the time taken by the man to cross the road is 4.1 seconds.

5 0

3 years ago