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3 years ago

If one half-life is 60 seconds, how old is the sample after 5 half-lives?

Physics

2 answers:

Serggg [28]3 years ago

6 0

Answer:

300 seconds?

Explanation:

yuradex [85]3 years ago

4 0

Answer: 300 years old

Explanation: 60x5=300

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Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$

Now

$F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf$

$F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf$

Stress is given by

$\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2$

The stress in the steel is $21678.47223\ lbf-in^2$

$\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2$

The stress in the steel is $383.1109\ lbf-in^2$

4 0

3 years ago

Shareen finds that when she drives her motorboat upstream she can travelwith a speed of only 8 m/s, while she moves with a speed

ANEK [815]

Let vb be the velocity of the motorboat and let vs be the velocity of the stream.

We know that when she drives upstream the velocity is 8 m/s, in this scenario the velocities point in opposite directions, then we have the equations:

$v_b-v_s=8$

When she drives downstream the velocites point in the same direction then we have the equation:

$v_b+v_s=12$

hence we have the system of equations:

$\begin{gathered} v_b-v_s=8 \\ v_b+v_s=12 \end{gathered}$

Solving the first equation for the velocity of the boat we have:

$v_b=8+v_s$

Plugging this in the second equation we have:

$\begin{gathered} 8+v_s+v_s=12 \\ 2v_s=4 \\ v_s=\frac{4}{2} \\ v_s=2 \end{gathered}$

Therefore, the velocity of the stream is 2 m/s

5 0

1 year ago

sound wave of frequency 640 Hz travels 8oom in 2.5 s. calculate: (a) speed of sound (b) wavelength of sound wave

kotegsom [21]

9.780s trying to add everything is pretty good it's stick around it always works for me

5 0

3 years ago

A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing

VARVARA [1.3K]

Explanation:

(a) The given data is as follows.

Length of the rod, L = 0.83 m

Mass of the rod, m = 110 g = 0.11 (as 1 kg = 1000 g)

At the lowest point, angular speed of the rod ( $\omega$ ) = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

I = $I_{CM} + mh^{2}$

= $\frac{1}{12}mL^{2} + m(\frac{L}{2})$

= $\frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}$

= 0.00631 + 0.415

= 0.42131 $kg m^{2}$

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

K = $\frac{1}{2}I \omega^{2}$

= $\frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}$

= 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b) According to the conservation of total mechanical energy of the rod, we have

$K_{i} + U_{i} = K_{f} + U_{f}$

$K_{i} = U_{f} - U_{i}$

or, mgh = K = 6.86 J

Therefore, h = $\frac{6.86}{mg}$

= $\frac{0.63}{0.11 \times 9.8}$

= 0.584 m

Hence, the center of mass rises 0.584 m far above that position.

6 0

3 years ago

Read 2 more answers

you are designing an airport for small planes. one kind of airplane that might use this airfield must reach a speed before takeo

denis23 [38]

(a)The required speed for take off will be 24.49 m/sec.

(b)The minimum length must the runway have 193 m.

<h3>What is speed?
</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its SI unit is m/sec.

(a)The required speed for take off will be 24.49 m/sec.

From the Newtons third equation of motion;

$\rm v^2 = u^2+2as \\\\ v=0+2 \times 2 \times 150 \\\\ v=\sqrt{600}\\\\ v=24.49 \ m/sec$