<h3><u>Answer;</u></h3>
C. Supersaturated
<h3><u>Explanation</u>;</h3>
- Solutions are homogeneous mixtures that are created by mixing a solute and a solvent. Solute is the substance present in smaller amounts that dissolves in a solvent such as water which is the substance present in larger amount.
- A solution, can be<u> unsaturated, saturated or supersaturated. An unsaturated solution</u> is a solution that contains less solute that can be dissolved, it doesn't contain the maximum amount of solute.
- <u>A saturated solution</u> is a solution containing the maximum amount of solute that can be dissolved at a given temperature. Any additional solute will remain undissolved in the container.
- <u>A supersaturated solution</u> is a solution created when a solution is carefully cooled because it contains more solute than the solubility allows.
Answer:
The dose is 6 mSV
Explanation:
The absorbed dose (in gray - Gy) is the amount of energy that ionizing radiation deposits per unit mass of tissue. That is,
Absorbed dose = Energy deposited / Mass
while Dose equivalent (DE) (in Seivert -Sv) is given by
DE = Absorbed dose × RBE (Relative biological effectiveness)
First, we will determine the Absorbed dose
From the question, Energy deposited = 30mJ and Mass = 50kg
From,
Absorbed dose = Energy deposited / Mass
Absorbed dose = 30mJ/50kg
Absorbed dose = 0.6 mGy
Now, for the Dose equivalent (DE)
DE = Absorbed dose × RBE
From the question, RBE = 10
Hence,
DE = 0.6mGy × 10
DE = 6 mSv
Answer:
x = 2,864 m
, Ra = 32.1 m
Explanation:
Let's solve this problem in parts, let's start by finding the intensity of the sound in each observer
observer A β = 64 db
β = 10 log Iₐ / I₀
where I₀ = 1 10⁻¹² W / m²
Iₐ = I₀ 10 (β/ 10)
let's calculate
Iₐ = 1 10⁻¹² (64/10)
Iₐ = 2.51 10⁻⁶ W / m²
Observer B β = 85 db
I_b = 1 10-12 10 (85/10)
I_b = 3.16 10⁻⁴ W / m²
now we use that the emitted power that is constant is the intensity over the area of the sphere where the sound is distributed
P = I A
therefore for the two observers
P = Ia Aa = Ib Ab
the area of a sphere is
A = 4π R²
we substitute
Ia 4pi Ra2 = Ib 4pi Rb2
Ia Ra2 = Ib Rb2
Let us call the distance from the observer be to the haughty R = ax, so the distance from the observer A to the haughty is R = 35 ax; we substitute
Ia (35 -x) 2 = Ib x2
we develop and solve
35-x = Ra (Ib / Ia) x
35 = [Ra (Ib / Ia) +1] x
x (11.22 +1) = 35
x = 35 / 12.22
x = 2,864 m
This is the distance of observer B
The distance from observer A
Ra = 35 - x
Ra = 35 - 2,864
Ra = 32.1 m
Answer:
Um okay? thank you for the points have a good say and be careful out there!
Explanation:
Explanation:
When m=<em>mass</em>
G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>
<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>
<em>M</em><em>g</em><em>h</em>
<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>
6×10×h
=60joules
