Answer:
A star is a huge sphere of very hot, glowing gas. Stars produce their own light and energy by a process called nuclear fusion. Fusion happens when lighter elements are forced to become heavier elements. When this happens, a tremendous amount of energy is created causing the star to heat up and shine.
Answer:
the potential energy of this body is 245 J.
Explanation:
Given;
mass of the body, m = 250 g = 0.25 kg
height from which the body was dropped, h = 100 m
acceleration due to gravity, g = 9.8 m/s²
The potential energy of this body is calculated as;
P.E = mgh
substitute the given values and solve for the potential energy of this body;
P.E = 0.25 x 9.8 x 100
P.E = 245 J.
Therefore, the potential energy of this body is 245 J.
Hydrogen = 1
oxygen = 8
but then remember the 2
hope that kinddaaa helps xoxo
Answer:
The correct answer is "".
Explanation:
According to the question,
The work will be:
⇒
Thus the above is the correct answer.
Answer: The workers may come only as far as 0.24m
<em>(Note: The queston is incomplete. Here is the complete question below)</em>
A shielded gamma ray source yields a dose rate of 0.048rad/h at a distance of 1.0m for an average-sized person. If workers are allowed a maximum dose of 5.0 rem in 1 year, how close to the source may they operate, assuming a 35-h work week? Assume that the intensity of radiation falls off as the square of the distance. (It actually falls off more rapidly than 1/r2 because of absorption in the air, so your answer will give a better-than-permissible value.)
Explanation:
Note: 1rem = 1rad
Dose rate per hour from a distance of 1.0m = 0.048rem/h
maximum daily dose in a year for a 35-h week = 5rem/yr / (35h*52*1yr) = 0.0027rem/h.
The required distance is obtained from the <em>inverse square law </em>for radiation<em> which states that the intensity of the radiation </em><em>(I)</em><em> decreases in proportion to the inverse of the distance from the source </em><em>(d)</em><em> squared.</em>
From the law, I₁/I₂ = d₁²/d₂²
Therefore, by knowing the intensity at one distance, one can find the intensity at any other distance.
From the given values I₁ = 0.048rem/h, I₂ = 0.0027rem/h, d₁ = 1.0m, d₂ = ?
d₂² = I₂d₁²/ I₁
d₂² = 0.0027rem/h * (1.0)²/ 0.048rem/h
d₂² = 0.05625m²
d₂ = √(0.05625)
d₂ = 0.24m