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3 years ago

6

Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou

chdown

2 answers:

ch4aika [34]3 years ago

6 0

Answer:

2s

Explanation:

Given parameters:

Distance = 30m

Speed = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

Speed = $\frac{disance }{time}$

Time taken = $\frac{distance }{speed }$ = $\frac{30}{15}$ = 2s

The time it takes is 2s

Anton [14]3 years ago

5 0

<u>Given:</u>

<u>30 meters</u>
<u>Speed of 15 m/s</u>

<u>We will solve:</u>

<u>Speed = </u> $\frac{distance}{time}$ <u></u>

<u></u>

<u> Time taken = </u> $\frac{distance}{time}$ <u>= </u> $\frac{30m}{15m/s}$ <u>= 2s</u>

<u></u>

<u>The time it takes is 2s</u>

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Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

$\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} }$ (1)

Where:

$n_{o}$ - Initial amount of carbon-14.

$n(t)$ - Current amount of carbon-14.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

Then, we clear the time within the formula:

$t = -\tau \cdot \ln \frac{n(t)}{n_{o}}$ (2)

In addition, time constant can be calculated by means of half-life of carbon-14 ( $t_{1/2}$ ), measured in years:

$\tau = \frac{t_{1/2}}{\ln 2}$

If we know that $\frac{n(t)}{n_{o}} = 0.25$ and $t_{1/2} = 5730\,yr$ , then the age of the organism is:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.25$

$t \approx 11460.001\,yr$

The age of the organism is approximately 11460 years.

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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

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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.

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