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2 years ago

Which statements describe the applications of nuclear medicine scans? Check all that apply.

Physics

2 answers:

coldgirl [10]2 years ago

4 0

Answer:

- asses disease progression and tissue function

- utilize a biologically active molecule

- utilize a radionuclide tracer

Explanation:

Mamont248 [21]2 years ago

4 0

Answer:

the answer is 1,3,5

Explanation:

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Two roads intersect at right angles, one going north-south, the other east-west. an observer stands on the road 60 meters south

Sloan [31]

observer is standing at distance d = 60 m south from the intersection

cyclist is travelling at speed v = 10 m/s

now after t = 8 s its displacement from intersection is given by

$x = 10*8 = 80 m$

so the position of cyclist makes an angle with the observer

$\theta = tan^{-1}\frac{80}{60} = 53 degree$

now the component of velocity of cyclist along the line joining its position with the observer is given as

$v = v_o cos\phi$

here

$\phi = 90 -\theta$

$\phi = 90 - 53 = 37 degree$

$v = 10 cos37 = 8 m/s$

so at this instant cyclist is moving away with speed 8 m/s

7 0

2 years ago

Read 2 more answers

An object is dropped from rest into a pit, and accelerates due to gravity at roughly 10 m/s2. It hits the ground in 5 seconds. A

vitfil [10]

Answer:

Second pit is 375 m deeper compared to first pit.

Explanation:

We have equation of motion s = ut + 0.5at²

First object hits the ground after 5 seconds,

Initial velocity, u = 0 m/s

Acceleration, a = 10 m/s²

Time, t = 5 s

Substituting,

s = ut + 0.5 at²

s = 0 x 5 + 0.5 x 10 x 5²

s = 125 m

Depth of pit 1 = 125 m

Second object hits the ground after 10 seconds,

Initial velocity, u = 0 m/s

Acceleration, a = 10 m/s²

Time, t = 10 s

Substituting,

s = ut + 0.5 at²

s = 0 x 10 + 0.5 x 10 x 10²

s = 500 m

Depth of pit 2 = 500 m

Difference in depths = 500 - 125 = 375 m

Second pit is 375 m deeper compared to first pit.

7 0

1 year ago

Match the measuring instrument with the appropriate statement.

yaroslaw [1]

<u>Answer:</u>

1A, 2E, 3C, 4F, 5B, 6D

<u>Explanation:</u>

Tilt meter is a an equipment used to measure very small changes from vertical level. So it is a correct match for blank space in statement A.

Richter Scale is a scale of measurement of earthquake strength, So it is a correct match for blank space in statement E.

Mercalli Intensity Scale measures effect of an earthquake, like damage caused.So it is a correct match for blank space in statement C.

Seismograph is earthquake wave measuring instrument. So it is a correct match for blank space in statement F.

Correlation Spectrometer (C.O.S.P.E.C.) is used to measure sulfur dioxide content in smoke. So it is a correct match for blank space in statement B.

Moment Magnitude Scale is an earthquake measuring scale for great earthquakes. So it is a correct match for blank space in statement D.

7 0

2 years ago

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of

adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0

1 year ago

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0

2 years ago