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3 years ago

A radiograph of the forearm is produced using 4 mA at 75 kVp. What kvp would be required to double the exposure?

Physics

1 answer:

BabaBlast [244]3 years ago

8 0

Answer:

Required KVP = 86 KVP (Approx)

Explanation:

Given:

Current KVP = 75 KVP

Find:

KVP required to double exposure

Computation:

According to 15% rule of KVP,

15% change increse in KVP required to get double exposure.

So,

Required KVP = Current KVP + Current KVP(15%)

Required KVP = 75 KVP + 75 (15%)

Required KVP = 75 KVP + 11.25 KVP

Required KVP = 86.25 KVP

Required KVP = 86 KVP (Approx)

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In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas

Sonja [21]

Answer:

The speed of disk is 1.98 $\frac{m}{s}$

Explanation:

Given:

Mass of $m = 0.099$ kg

Spring constant $k = 244 \frac{N}{m}$

Compression of spring $x = 4 \times 10^{-2}$ m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

$v = \sqrt{\frac{k x^{2} }{m} }$

$v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }$

$v = 1.98 \frac{m}{s}$

Therefore, the speed of disk is 1.98 $\frac{m}{s}$

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3 years ago

One consequence of Newton's third law of motion is that __________. A. Every object that has mass has inertia B. A force acting

vovangra [49]

Answer:

Explanation:

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3 years ago

Which of these help create radio waves?

inysia [295]

The one that help create radio waves is :
Changing electric and magnetic fields applied at right angles

Radio waves are transverse wave, which means that the oscillations occurring perpendicular to the direction of energy transfer

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3 years ago

The cardiovascular system consists of

guapka [62]

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3 years ago

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago