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3 years ago

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You are working with a 3 T MRI scanner with maximum gradient strength of 40 mT/m in all 3 axes (x,y,z). Your acquisition uses ma

ximum gradient strength to minimize the duration of the data acquisition window. When the Gx is the only gradient that is turned on, what is the rotation frequency for spins located at x=5cm, y=5cm? The gyromagnetic ratio is γ=42.576 MHz/T

1 answer:

inn [45]3 years ago

7 0

Answer:

Rotational frequency=ω/2 $\pi$

Explanation:

Hence rotational frequency will be = 13.55 kHz

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Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip’s speed at impact

Westkost [7]

Answer:

-5.8868501529 m/s² or -5.8868501529g

0.118909090909 s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{1.3^2-2^2}{2\times 0.02}\\\Rightarrow a=-5.8868501529\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g$

The acceleration is -5.8868501529 m/s² or -5.8868501529g

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{1.3-2}{-5.8868501529}\\\Rightarrow t=0.118909090909\ s$

The time taken is 0.118909090909 s

7 0

3 years ago

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

Margarita [4]

Answer:

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

Explanation:

We have to make a hollow sphere of inner radius $r_1$ and outer radius $r_2$ .

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

$V_t=\frac{4}{3} \pi.r_2^3$

And the volume of the hollow space in the sphere:

$V_h=\frac{4}{3} \pi.r_1^3$

Therefore the net volume of material required to make the sphere:

$V=V_t-V_h$

$V=\frac{4}{3} \pi(r_2^3-r_1^3)$

Now let the density of the of the material be $\rho$ .

<u>Then the mass of the material used is:</u>

$m=\rho.V$

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

4 0

4 years ago

Are sound waves transverse waves or longitudinal waves? Why?

aleksandrvk [35]

Answer: Longitudinal Waves

Explanation: The particles of the medium pass parallel to the course that the heart beat moves. This sort of wave is a longitudinal wave. Longitudinal waves are usually characterised through particle motion being parallel to wave movement. A valid wave traveling thru air is a classic example of a longitudinal wave.

4 0

4 years ago

Read 2 more answers

A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the lad

pantera1 [17]

Answer: 45.3°

Explanation:

Given,

Length of ladder = l

Weight of ladder = w

Coefficient of friction = μs = 0.495

Smallest angle the ladder makes = θ

If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,

N1 = mg and

N2 = μmg

Moment at a point A in the clockwise direction is

N2 Lsinθ - mg.(L/2).cosθ = 0

μmgLsinθ - mg.(L/2).cosθ = 0

μmgLsinθ = mg.(L/2).cosθ

μsinθ = cosθ/2

sin θ / cos θ = 1 / 2μ

Tan θ = 1 / 2μ

Substituting the value of μ = 0.495, we have

Tan θ = 1 / 2 * 0.495

Tan θ = 1 / 0.99

Tan θ = 1.01

θ = tan^-1(1.01)

θ = 45.3°

3 0

3 years ago

Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe

RideAnS [48]

Sarah's acceleration is $-0.49 m/s^2$

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

$F_f = \mu mg$

where

$\mu$ is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

$F=ma$

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

$F=-\mu mg = ma$

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

$a=-\mu g$

And substituting the following values:

$\mu = 0.05$ (coefficient of friction)

$g=9.81 m/s^2$ (acceleration of gravity)

we find:

$a=-(0.05)(9.81)=-0.49 m/s^2$

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4 0

3 years ago