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3 years ago

Which form of electromagnetic radiation with a relatively high amount of energy is used to sanitize lab goggles?

Physics

1 answer:

alexdok [17]3 years ago

5 0

Answer:

Gamma radiation.

Explanation:

Electromagnetic waves are waves produced with respect to the interaction of electric field and magnetic field. They can be arranged with respect to their increasing or decreasing wavelength to form a pattern called electromagnetic spectrum.

Gamma radiation can be used for sterilization purpose due to the fact that it has a relatively high amount of energy, and do not produce any characteristic radiation.

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Will mark brainliest

sweet-ann [11.9K]

Answer: Wouldn't it just be her blocks all walked in an hour added together?

Explanation: 5+2+3+2=12 so 12 blocks an hour?

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3 years ago

After a big snowfall, you take your favorite rocket‑powered sled out to a wide field. The field is 223 m223 m across, and you kn

pogonyaev

Answer:

11.7 s

Explanation:

In this problem, the rocket is moving in a uniform accelerated motion. We have the following data:

d = 223 m, the distance that the sled has to cover

$a=3.25 m/s^2$ , the acceleration of the rocket

We can use therefore the following SUVAT equation:

$d=ut+\frac{1}{2}at^2$

where

d is the distance

u = 0 is the initial velocity of the sled (it starts from rest)

t is the time

a is the acceleration

Re-arranging the equation and substituting the numbers, we find the time it takes for the rocket to cross the field:

$d=\frac{1}{2}at^2\\t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(223)}{3.25}}=11.7 s$

6 0

3 years ago

What is the momentum of an 18-kg object moving at 0.5 m/s?

Arada [10]

Answer:

9kgm per second

Explanation:

3 0

2 years ago

Read 2 more answers

The perimeter of a sector of a circle is the sum of the two sides formed by the radii and the length of the included arc. A sect

snow_tiger [21]

Answer:

Length of the arc of this sector, l = 14 cm

Explanation:

It is given that, the perimeter of a sector of a circle is the sum of the two sides formed by the radii and the length of the included arc.

Perimeter of sector, P = 28 cm

Area of sector, $A=49\ cm^2$

According to figure,

2r + l = 28 ............(1)

Area of sector, $A=\dfrac{\theta}{360}\times \pi r^2$

Where, $\theta$ is in radian and $\theta=\dfrac{l}{r}$

Since, $1^{\circ}=\dfrac{\pi}{180}\ radian$

$A=\dfrac{l}{2\pi r}\times \pi r^2$

$r=\dfrac{98}{l}$

Put the value of r in equation (1) so,

$2\times (\dfrac{98}{l})+l=28$

$l^2-28l+196=0$

On solving above equation for l we get, l = 14 cm. So, the length of the arc of this sector is 14 cm. Hence, this is the required solution.

5 0

3 years ago

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

lyudmila [28]

Answer:

d) $7.94\times 10^{9}$

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

$\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$12 = log\left ( \frac{I_{1}}{10^{-12}} \right )$ Eq-1

sound level due to whisper is given as

$\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )$ Eq-2

subtracting Eq-2 from Eq-1

$12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )$

$9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )$

$\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}$

6 0

3 years ago