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4 years ago

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What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 85.0% absorbed, puts 470 J of energy

into a circular spot 2.60 mm in diameter in 4.00 s?

1 answer:

Advocard [28]4 years ago

8 0

Answer:

26036485.6433 W/m²

Explanation:

E= Energy = 470 J

t = Time = 4 seconds

d = Diameter = 2.6 mm

Power is given by

$P=\dfrac{E}{t}$

Intensity is given by

$I=\dfrac{P}{\pi r^2}\\\Rightarrow 0.85I=\dfrac{E}{t\dfrac{\pi}{4} d^2}\\\Rightarrow I=\dfrac{470}{0.85\times 4\times \dfrac{\pi}{4}\times (2.6\times 10^{-3})^2}\\\Rightarrow I=26036485.6433\ W/m^2$

The intensity of the laser beam is 26036485.6433 W/m²

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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

So

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

B)

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

C)

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

D)

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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4 years ago

Sleep is a component of which part of the health triangle

katrin [286]

Answer:

physical health

Explanation:

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3 years ago

What do conduction, radiation, and convection work together to do?

Semmy [17]

Well, conduction is the passing of heat by touch. Convection is the passing along of heat by movement (e.g. hot air rising from a radiator to heat the rest of the room), and radiation is the heating of something through electromagnetic rays without any contact. For example, the earth absorbs the heat radiated from the son (radiation). The air that touches the earth gets warmed up (conduction), and rises, and as it rises, it passes its heat along to more air around it (convection).

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4 years ago

I’m stuck on this problem, help is very appreciated!!

yaroslaw [1]

Explanation:

Use the equation F=m×a

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3 years ago

Read 2 more answers

1. Is the sequence geometric? If so, identify the common ratio.

frozen [14]

1) <span>yes;2 6*2=12 12*2=24 24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1
</span>
a = first term, i.e. 5
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
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<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
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6 0

4 years ago