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2 years ago

What is Newton's 3rd law of motion?

Physics

1 answer:

eimsori [14]2 years ago

8 0

Answer:

Explanation:

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Which applications, either for diagnostic purposes or for therapeutic purposes, involve the use of X-rays

Alina [70]

The applications, either for diagnostic purposes or for therapeutic purposes, involve the use of X-rays----CT scan radiography ,external beam radiation therapy, fluoroscopy.

How is CT used for treatment planning?

CT planning enables more accurate localisation of both tumour and normal organs in addition to providing an accurate body contour and inhomogeneity corrections.

What is difference between CT scan and fluoroscopy?

Overall, fluoroscopy is a safe procedure, but potential risks include burns or radiation-induced injuries to the skin. On the other hand, CT scans are still snapshots of a “slice” of the body. They use X-rays to help your doctor view important organs

What is a fluoroscopic procedure?

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Learn more about diagnosis:

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6 0

1 year ago

A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat

Nostrana [21]

Answer:

Approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ (assuming that the boiling point of water in this experiment is $100\; \rm ^\circ C\!$ .)

Explanation:

Latent heat of condensation/evaporation of water: $2260\; \rm J \cdot g^{-1}$ .

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to $\rm J \cdot g^{-1}$ .

Specific heat of water: $4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}$ .

Specific heat of copper: $0.39\; \rm J \cdot g^{-1}\cdot K^{-1}$ .

The temperature of this calorimeter and the $250\; \rm g$ of water that it initially contains increased from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ . Calculate the amount of energy that would be absorbed:

$\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}$ .

$\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}$ .

Hence, it would take an extra $585\; \rm J + 31500\; \rm J = 32085\; \rm J$ of energy to increase the temperature of the calorimeter and the $250\; \rm g$ of water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

Assume that it would take $x$ grams of steam at $100\; \rm ^\circ C$ ensure that the equilibrium temperature of the system is $50\; \rm ^\circ C$ .

In other words, $x\; \rm g$ of steam at $100\; \rm ^\circ C$ would need to release $32085\; \rm J$ as it condenses (releases latent heat) and cools down to $50\; \rm ^\circ C$ .

Latent heat of condensation from $x\; \rm g$ of steam: $2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J$ .

Energy released when that $x\; {\rm g}$ of water from the steam cools down from $100\; \rm ^\circ C$ to $50\; \rm ^\circ C$ :

$\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}$ .

These two parts of energy should add up to $32085\; \rm J$ . That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

$(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J$ .

Solve for $x$ :

$x \approx 13$ .

Hence, it would take approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ for the equilibrium temperature of the system to be $50\; \rm ^\circ C$ .

4 0

2 years ago

A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between

Ulleksa [173]

Answer:

90x20=1800

Explanation:

just multiply 10 & 9 and then mutiply 90x20 or 20x90

4 0

2 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

What type of motion occurs when an object spends around and axis without altering its linear position?

ch4aika [34]

I believe it is called centripetal force

7 0

3 years ago