This is because The 15% rule states that changing the kVp by 15% has the same effect as doubling the mAs, or reducing the mAs by 50%; that is A 15% increase in kVp has the same effect as doubling the mAs and vice versa but in this case reducing mas causes more Compton scatter because Compton scattering occurs at a lower kvp

8 0

3 years ago

Starting from zero, an electric current is established in a circuit made of a battery of emf E, a resistor of resistance R and a

igor_vitrenko [27]

Answer:

time constant will decrease and steady state current will decrease on increasing the resistance

Explanation:

As we know that the EMF of cell is E which is used to connected across a resistor and an inductor.

So we will have

$E - iR - L\frac{di}{dt} = 0$

here we know that

$i = \frac{E}{R}(1 - e^{-Rt/L})$

now here we have

$\tau = \frac{L}{R}$

so if we increase the value of resistance of the wire then the time constant will decrease

and hence it will take less time to reach near the steady state value

also the steady state current will be smaller in that case

8 0

3 years ago

A 50.6 g ball of copper has a net charge of 1.6 µc. what fraction of the copper's electrons have been removed? (each copper atom

Fynjy0 [20]

First figure out how many atoms you have with Avogadro's number. Since there are 63.5 grams/mol and you have 50.6 grams, you have (50.6/63.5)6.022E23=4.7986E23 atoms. Since there are 29 protons per atom, there are also 29 electrons per atom, so you should have a total of
29*4.7986E23=1.3916E25 electrons.
Since there is a positive charge you know some of these electrons are missing. How many are missing can be found by dividing the charge you have by the charge on the electron: 1.6E-6/1.6022E-19 = 9.98627E12 electrons are missing.
Now take the ratio of what is missing to what there should be:
9.98627E12/1.3916E25 = 7.1760873E-13

5 0

3 years ago