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makvit [3.9K]
3 years ago
7

Help please number 8 an 9

Physics
1 answer:
Murrr4er [49]3 years ago
5 0
8) The reason is that MRIs use radio-waves, which are electromagnetic waves with lower frequency, and so lower energy, than X-rays.
In fact, typical frequencies for radio waves are f=10-10^9 Hz, while X-rays have typical frequencies of 10^{16} - 10^{19} Hz. Since the energy of a wave is given by
E=hf
where h is the Planck constant and f the frequency, we see that the lower the frequency, the lower the energy. Therefore, radiowaves have less energy than X-rays, so they are less ionising and less dangerous than X-rays.

9) The false statement is:
a) gamma-rays are low-frequency waves.
It's actually the opposite, in fact: gamma rays are the electromagnetic waves with higher frequencies (above 10^{20}Hz).

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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
changes to OSHA's regulations in 2013 require chemical information be provided via the _______ format. a) safety data sheets b)
Talja [164]

Don't listen to the other guy I just took the test and got it wrong because of him..

I re-took it and the correct answer is

A) Safety Data Sheets (SDS)

8 0
3 years ago
A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi
Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0
3 years ago
A circular loop of wire with radius r=0.0250 m and resistance r=0.390 ohms is in a region of spatially uniform magnetic field. t
slavikrds [6]

Answer:

0.0133 A

Explanation:

The time at which B=1.33 T is given by  

1.33 = 0.38*t^3  

t = (1.33/0.38)^(1/3) = 1.52 s  

Using Faraday's Law, we have  

emf = - dΦ/dt = - A dB/dt = - A d/dt ( 0.380 t^3 )  

Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²

emf = - A*(3*0.38)*t^2  

thus, the emf at t=1.52 s is  

emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V  

if the resistance is 0.390 ohms, then the current is given by  

I = V/R = 0.0052/0.390 = 0.0133 A

3 0
3 years ago
Need help! Thanks for whoever answers.
Hunter-Best [27]

Answer:

2021 + 100000000

Explanation:

i o g kk ybb

5 0
3 years ago
Read 2 more answers
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