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Luda [366]
3 years ago
15

In an effort to decrease the mAs of an exposure, the 15% rule of kVp change may be considered. Changing the original kVp of 84 u

sing the 15% rule will have what impact
Physics
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

Will result in :Greater Compton scatter interaction

Explanation

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

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Can somebody help please !<br><br> a. -8.3 m/s<br> b.-4.2 m/s<br> c.-0.12 m/s<br> d. 0 m/s
Gnom [1K]
The answer is a
the \: answer \: is \: a.
4 0
3 years ago
Scientists study how the continents move. Why might scientists use a model
tigry1 [53]

Answer:

B. It is too slow to observe directly

Explanation:

They move too slow to be able to observe how they move.


I hope it helps! Have a great day!
bren~

3 0
2 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

4 0
3 years ago
Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,
yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

     = (9×10^9)×(q)/(24.0)

     = 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

                       R^2 = 9×r^2

E2 = k×q/(9×r^2)

     = k×q/(9×375000000q)

     = k/(9×375000000)

     = (9×10^9)/(9×375000000)

     = 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0
3 years ago
Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various
Alexxx [7]

Answer:

Q_{2}=1200cm^{3}/s

Explanation:

Given data

Q₁=200cm³/s

We know that:

F=n\frac{vA}{l}\\

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

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3 0
3 years ago
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