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1 year ago

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Any ferrous metal object within or near the mri magnet has the potential of becoming a projectile. this is commonly referred to

as a ___ injury/effect

1 answer:

PtichkaEL [24]1 year ago

4 0

This phenomena is referred to as metal projectile injury/effect. MRI stands for magnetic resonance imaging. As expected, your body is subjected to a magnetic field. This is a technique to look inside your body without cutting it open. Metals should not be placed anywhere near an MRI because it will cause the metals to follow the direction of the magnetic field.

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Two identical ladders are 3.0 m long and weigh 600 N each. They are connected by a hinge at the top and are held together by a h

ruslelena [56]

Answer:

The tension in the rope is 281.60 N.

Explanation:

Given that,

Length = 3.0 m

Weight = 600 N

Distance = 1.0 m

Angle = 60°

Consider half of the ladder,

let tension be T, normal reaction force at ground be F, vertical reaction at top hinge be Y and horizontal reaction force be X.

$Y+F=600$ ....(I)

$X=T$ .....(II)

On taking moment about base

$X\times l\cos\theta+Y\times l\sin\theta-F\dfrac{l}{2}\sin\theta-T\times d=0$

Put the value into the formula

$X\times3\cos30+Y\times3\sin30-600\times1.5\sin30-T\times1=0$

$3\cos30 T-T=600\times1.5\sin30-Y \times3\sin30$

$1.598T=450-1.5(600-F)$ ....(III)

We need to calculate the force for ladder

$2F=600\trimes 2$

$F=600\ N$

We need to calculate the tension in the rope

From equation (3)

$1.598T=450-1.5(600-600)$

$1.598T=450$

$T=\dfrac{450}{1.598}$

$T=281.60\ N$

Hence, The tension in the rope is 281.60 N.

7 0

2 years ago

Read 2 more answers

A tightly sealed glass jar is an example of which type of system?

noname [10]

D. hope it helps. :D

4 0

2 years ago

Read 2 more answers

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers

A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t

azamat

Answer:

(a) $x'(t)= 142$

(b) 142

(c) $y'(t)= -32t+5$

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$ , it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

(c) $y(t) = - 16t^2+5t+5$

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$ .

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$ .

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$

6 0

2 years ago

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat

Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0

2 years ago