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3 years ago

Which two statements describe how ultrasound technology produces an image of a baby before it is born?

Physics

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:a and c

Explanation:

garik1379 [7]3 years ago

7 0

Answer:

C,D

Explanation:

an image is created based on the amont of time it takes for a sound wave to return

Tissues in the babys body reflect high-frequency sound waves

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How long does it take a man to travel 6 km if his speed is 3km/h?

Ymorist [56]

why did my answer get deleted??

oh yeah i put a link on there- oopsies.

I wont this time!

I got 30!

3 0

2 years ago

At which point does the planet have the least gravitational force acting on it?

Elza [17]

Answer:

At which point does the planet have the least gravitational force acting on it?

Explanation:

In an elliptical orbit, when a planet is at its furthest point from the Sun, it is under the least amount of gravity, meaning that the force of gravity is strongest when it is closest.

5 0

3 years ago

Read 2 more answers

Indicate whether a change in the value of each of the following determinants of demand leads to a movement along the demand curv

Inessa05 [86]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Change in market price is m<span>ovement along the demand curve. </span>

4 0

3 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. T

gavmur [86]

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

I = 5mℓ²

4 0

3 years ago