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2 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]2 years ago

8 0

Answer:a and c

Explanation:

garik1379 [7]2 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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Which of these diagrams most accurately illustrates what happens when a ray of light from underwater strikes the smooth surface

laiz [17]

Diagram D!!!!!!!!!!!!!!!

4 0

2 years ago

Read 2 more answers

Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force const

IRISSAK [1]

First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:

$\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}$

Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:

$\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}$

Therefore the speed is 1.026 m/s.

5 0

10 months ago

To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p

il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0

2 years ago

A centrifuge rotates so that 0. 25 m from the center is traveling at 343 m/s (the speed of sound). How many RPM is this? What is

andrew-mc [135]

Answer:

[13,101 RPM]

[1372 rad/s]

Explanation:

8 0

2 years ago

g is incident on 3 successive sheets of polarizing material. The transmission axis of the first sheet is vertical. The transmiss

murzikaleks [220]

Answer:

The intensity of light passing from the third polarizer is 3Io/16.

Explanation:

The law of Malus is given by

$I=I_o cos^2\theta$

Let the incident intensity of light is Io.

The intensity of light passing from the first polarizer is

$I' = \frac{I_o}{2}$

The intensity of light passing from the second polarizer is

$I''=\frac{I_o}{2}\times cos^230 =\frac{3I_o}{8}$

The intensity of light passing from the third polarizer is

$I''' = \frac{3I_o}{8}\times cos^2 60\\\\\\I''' = \frac{3I_o}{16}$

6 0

3 years ago