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3 years ago

14

Part One–Research

1 answer:

Nimfa-mama [501]3 years ago

7 0

Answer: i need help to lpl

Explanation:

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with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d

Lubov Fominskaja [6]

The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.

<h3>How does kVp impact the exposure to digital receptors?</h3>

The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.

<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>

Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.

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5 0

1 year ago

Explain how use of EGR is effective in reducing NOx emissions 4. In most locations throughout the U.S., the octane number of reg

TiliK225 [7]

Answer:please see attached file

Explanation:

3 0

4 years ago

A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu

krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

$\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}$

$10*2 = \sqrt{1 + 8f^2 - 1$

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

$\frac{V_1}{\sqrt{g*y_1}} = 7.416$

$V_1 = 7.416 *\sqrt{32.2*1}$

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

$V_2 = \frac{3666.66}{80*10}$

= 4.208 ft/sec

Froude number, F2 =

$\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}$

F2 = 0.235

d) $El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}$

$El = \frac{(10-1)^3}{4*1*10}$

$= \frac{9^3}{40}$

= 18.225ft

e) for critical depth, we use :

$y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3$

= 3.80 ft

7 0

3 years ago

Read 2 more answers

Water enters a boiler at a temperature of 110°F. The boiler is to produce 2000 lb/hr of steam at a pressure of 130 psia. How man

strojnjashka [21]

Yes it does the answer is no

5 0

2 years ago

Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi

lawyer [7]

Answer:

$\vec F_{A} = -67500\,N\cdot (i + j)$

Explanation:

The position of each point are the following:

$A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)$

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

$\vec F_{A} = \vec F_{AB} + \vec F_{AC}$

$\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j$

$\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)$

$\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)$

$\vec F_{A} = -67500\,N\cdot (i + j)$

6 0

3 years ago