Ask question

Ask question

All categories

1 year ago

12

When an MRI machine are use to capturing images of a person's body the produce extremely powerful magnetic fields what might be

some issues that an MRI technician would need to work on around due to these magnetic fields

1 answer:

aliina [53]1 year ago

4 0

Answer:

Technique to reduce Noise, magnetic attraction and Twitching sensation

Explanation:

Changing magnetic fields cause loud knocking noise this can be overcome by ear protection.

Strong and Static magnetic fields attract magnetic objects that can be avoided by screening of people and objects before entering in that area.

Twitching Sensation due to nerve stimulation can be avoided by reducing the exposure to magnetic field.

Please mark branliest if you are satisfied with the answer. Thanking you in anticipation.

You might be interested in

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

2 years ago

Why are you more likely to get a concussion when hit by a field hockey ball than a

Aleksandr-060686 [28]

Answer:

because it can be hard

Explanation:

I said that because they be on bed rest

6 0

3 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

7 0

2 years ago

A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle

Brilliant_brown [7]

Answer:

$F_2 = 1.10 \mu N$

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

$F = \frac{kq_1q_2}{r^2}$

now since it depends inverse on the square of the distance so we can say

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

now we know that

$r_2 = 18.2 mm$

$r_1 = 12.2 mm$

also we know that

$F_1 = 2.45 \mu N$

now from above equation we have

$F_2 = \frac{r_1^2}{r_2^2} F_1$

$F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)$

$F_2 = 1.10 \mu N$

5 0

3 years ago

If you have 5.112 grams of ammonia (NH3), then how many molecules of ammonia do you have?

Dominik [7]

I hope it helps! good luck in chem!

7 0

2 years ago