Which of the following statements about features on the sun is true

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

Advocard [28]

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

8 0

4 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

When must scientific theories be changed

madreJ [45]

Answer:

when new information disproving the current theory becomes available.

Explanation:

hope this helps..

4 0

2 years ago

Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements

masya89 [10]

Answer:

When the starting and ending points are the same, the total work is zero.

Explanation:

option ( D )correct

A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.

8 0

3 years ago

Which of the following is most closely related to an activated complex?

Anon25 [30]

Among the choices above, the one that is most closely related to an activated complex is the transition state. The answer is letter D. This formation forms quickly and does not stay in a way compound is. It usually forms during the enzyme – substrate reaction.

5 0

3 years ago