Ask question

Ask question

All categories

3 years ago

8

What are some of the uses of ultra-sound

2 answers:

saw5 [17]3 years ago

7 0

Answer:

Ultrasound can be used for various purposes like diagnosing conditions, including those in the heart, blood vessels, liver and more. Yet most commonly to monitor growth and development of a fetus. A echocardiogram is a ultrasound of the heart.

Explanation:

kirill115 [55]3 years ago

3 0

Answer:

Ultrasound has been used in a variety of clinical settings, including obstetrics and gynecology, cardiology and cancer detection. The main advantage of ultrasound is that certain structures can be observed without using radiation. Ultrasound can also be done much faster than X-rays or other radiographic techniques

You might be interested in

Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.

Snezhnost [94]

The average velocity can be calculated using the formula:

v = d / t

For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

Therefore the train has a constant acceleration of about 0.23 meters per second.

5 0

3 years ago

The sleigh is being pulled with a force of 800N and has a mass of 200 Kg.

kiruha [24]

The answer is 4 m/s/s

4 0

3 years ago

Read 2 more answers

Imvu?<br> ..............................

Dominik [7]

Imvu? Like the app?

4 0

3 years ago

Read 2 more answers

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago

Cavity wall insulation costs £240 but will save you £32 each year. What is the payback time for cavity

likoan [24]

Answer:

Thus, the payback time for cavity wall insulation is 7.5 years

5 0

2 years ago