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2 years ago

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Projectile effects are a hazard in __________________. magnetic resonance imaging (MRI) fields pediatric units emesis stations o

perating theaters

1 answer:

zhenek [66]2 years ago

3 0

Projectile effects are a hazard in MRI fields.

<h3>What is MRI field?</h3>

MRI stands for - Magnetic resonance imaging and it is a medical imaging technique that uses a magnetic field and radio waves to create detailed images of the organs and tissues in your body.

Projectile effects are a hazard in MRI fields, due to attraction exerted by the static magnetic field of the MRI scanner on ferromagnetic objects accidentally introduced into the MRI-scanner room.

Thus, Projectile effects are a hazard in MRI fields.

Learn more about MRI fields here: brainly.com/question/23730902

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Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called <u>Electrostatic</u> <u>Force</u>.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

2 years ago

Which scientific law states the relationship between an object's mass, acceleration, and amount of force acting on it? (3 points

Kisachek [45]

Answer:

4.Newton's second law of motion

4 0

2 years ago

Read 2 more answers

A supply bag is dropped from a rescue plane. After the bag falls for 3.2 seconds , what is the velocity of the bag?

loris [4]

Answer: -31.36 m/s

Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

$V_{f}=V_{o}+a.t$ (1)

Where:

$V_{f}$ is the final velocity of the supply bag

$V_{o}=0$ is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

$a=g=-9.8m/s^{2}$ is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

$t=3.2s$ is the time

Knowing this, let's solve (1):

$V_{f}=0+(-9.8m/s^{2})(3.2s)$ (2)

Finally:

$V_{f}=-31.36m/s$ Note the negative sign is because the direction of the bag is downwards as well.

8 0

2 years ago

If someone looks far enough into space, they should be able to see the beginning of the universe true of false

kodGreya [7K]

The answer is no, it would be impossible to see the beginning of the universe

8 0

2 years ago

****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)

notka56 [123]

but I think it's a and f

and

b and e

3 0

2 years ago