Answer:
b. 12.5 mAs, 70 kVp
Explanation:
The given parameter are;
The initial exposure factors := 10 mAs and 70 kVp
The initial Grid Ratio, G.R.₁ = 8:1
The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂ = 12:1
Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;
The GCF for G.R. 8:1 = 4
The GCF for G.R. 12:1 = 5
Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;
mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs
Therefore the new exposure factors are;
12.5 mAs, 70 kVp
Answer:
14 m/s
Explanation:
The following data were obtained from the question:
Mass = 50 kg
Initial velocity (u) = 0 m/s
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?
The velocity (v) with which the person hit the water can be obtained as shown below:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 10)
v² = 0 + 196
v² = 196
Take the square root of both side
v = √196
v = 14 m/s
Therefore, he will hit the water with a speed of 14 m/s
Answer:
<h3>The answer is 36 J</h3>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 3 × 12
We have the final answer as
<h3>36 J</h3>
Hope this helps you
Based on the equation KE = 1/2(m)(v^2), Kinetic Energy can be measured based on velocity. If an object has a large velocity, it have a larger kinetic energy than if the velocity is small.
Hope this helps.
If this helped you, please vote me as brainliest!
The mind boggles at the spectacle ... a 150-kg lady weighs about 330 pounds, and has the courage and the audacity to not only go to the beach, but to go out and float in the water, hoping that there are no wise little kids around, getting ready to yell to each other "Thar she blows, Cap'n !".
But we're only here to do the math, whether or not it has any physical significance in the real world.
This problem is really not that mysterious.
Since the star of this unlikely scenario is just barely floating, her weight must be equal to the weight of an equal volume of water. In other words, her average density (arrghhh, I can't take much more of this) is equal to the density of water.
Officially, 1 kg is the mass of fresh water that occupies a volume of 1 liter, so this relationship also very nearly describes the relationship between our heroine's mass and volume.
Her volume is very nearly 150 liters, or 0.15 cubic meter, and no further rounding is necessary, in more ways than one.
I would stay and convert that to some more familiar 'customary' units for you, honest I would. But I've been quite overcome by my visualizations of this whole matter, and I really must make a quick trip to the rail, immediately.
