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3 years ago

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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

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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A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

Readme [11.4K]

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$

$F_{t}-m.g=m.a$

$18.8-9.8m=5.22m$

$15.02m=18.8$

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: $F_{g}=m.g$ .

The weight for the hanging block is

$F_{g}=1.25*9.8$

$F_{g}=$ 12.25 N

Comparing tension and weight:

$\frac{12.25}{18.8}$ ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

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3 years ago

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un

Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within 1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

$B_{i} =\frac{hr/2}{k}$

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

$\pi$ Dh(T-Tinf)= $I^{2} R_{e}$

make T the subject of the equation , we have

T=25+ $\frac{100^2*0.01}{\pi*0.001*500 }$

T=88.7 degC

rate of chnage in temperature

dT/dt= $\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)$

at t=o and integrating both sides $\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}$

we have

$\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}$

t=8.31s

steady state temperature =88.7deg C

t=time within 1 degC of it steady stae is 8.31s

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3 years ago

In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of fric

lorasvet [3.4K]

Answer:

B. The maximum angle decreases

Explanation:

If θ be the maximum angle of a slope that allows a crate placed on it to remain at rest , following condition exists .

tanθ = μ , θ is called angle of repose . μ is coefficient of static friction .

So the tan of angle of repose θ is proportional to coefficient of static friction.

If coefficient of static friction is less than .7 , naturally angle of repose will also become less ,ie, it at lower angle of inclination , the object will start slipping .

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3 years ago

What is the full meaning of (i.p.s.m.n)

Anna11 [10]

Answer:

Intraductal Papillary Mucinous Neoplasm

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3 years ago

A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Yuri [45]

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.

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3 years ago

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