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2 years ago

Superman used X-ray vision to detect objects. Aside from a source of X-rays, what else would be needed for X-ray vision?

Physics

1 answer:

Arada [10]2 years ago

3 0

Answer:

an absorber of x-ray

Explanation:

To make x-ray detection/vision work, you will need at least two items: a source of x-ray and absorbed or x-ray.

The object you want to see itself doesn't have to be the source, but it has to absorb some of the rays instead. When doing a chest x-ray test, the medical employee will put your chest between absorber and source. The heart can absorb some of the rays so it will appear more white than lungs who made of air and won't absorb the rays.

You might be interested in

According to the article, which pattern of brain waves are most conducive to studying new information?

sashaice [31]

Alpha brain waves are those most conducive to studying new information.

When consciously alert, we generally function along a beta brain rhythm. In diminishing this rhythm to alpha, we transition into a state of physical and mental relaxation that is ideal for learning new information and storing facts and data. Studies have shown that the effect of decreasing brain rhythm is linked to feelings of increased mental clarity and remembrance. As it is a prime condition for synthetic thought and creativity, it becomes easier to visualize and create associations (information is better learned and absorbed by using such study methods).

Hope this helps! :)

7 0

2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

ioda

Answer:

$6.1875\times 10^{-8}$

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies $I=\frac {k}{r^{2}}$ where r is the distance

Since intensity is given then when at 3 m

$1.1\times 10^{-7}= \frac {k}{3^{2}}$

$k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}$

Since we have the constant then at 4m

Intensity, $I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}$

8 0

2 years ago

Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate

Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

$E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J$

Time required to burn the fat

$t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s$

Distance traveled by the bird

$s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m$

The bird will fly 455165.278 m

4 0

2 years ago

Read 2 more answers

Suppose you see two main-sequence stars of the exact same spectral type. Star 1 is dimmer in apparent brightness than Star 2 by

katrin2010 [14]

Options:

A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.

B. Star 1 is 100 times more distant than Star 2.

C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.

D. Star 1 is 10 times more distant than Star 2.

E. Star 1 is 100 times nearer than Star 2.

Answer:

D. Star 1 is 10 times more distant than star 2

Explanation:

For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.

Luminosity, L = 1/r²

Where r is the distance of the star to the earth

Since star 1 is dimmer in brightness than star 2 by a factor of 100,

L₁/L₂ = 1/100

i.e. L₁ = 1, L₂=100

L₁ = 1/r₁² ............(1)

1 = 1/r₁²

L₂ = 1/r₂²

100 = 1/r₂² .........(2)

divide equation (2) by equation (1)

100/1 = ( 1/r₂² )/ (1/r₁²)

100 = (r₁/r₂)²

r₁/r₂ = √100

r₁/r₂ = 10

r₁ = 10r₂

3 0

2 years ago