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3 years ago

Which of the following is not an application of Doppler technology?

2 answers:

8 0

The correct answer to the question above is The third Option: C; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology.

Hope this helps! :)

Reil [10]3 years ago

4 0

Answer:

ultrasound imaging of the liver.

Explanation:

Liver ultrasound is not an application of the Doppler effect. The use of ultrasound is more common in medicine, for example, during pregnancy.

Another application in medicine is with the movement of blood. It is a flowmeter which allows the blood flow to be discovered through the Doppler effect, since a sound wave affects the red blood cells that are in motion inside the vessel and collect the echo signal.

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A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A pie

liubo4ka [24]

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

Dielectric constant k=3.55.

a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

E=1.1mJ

b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.

6 0

3 years ago

a projectile is lunched with an initial speed of 60.0mm/s at an angle of 30.0° above the horizontal.The projectile lands on a hi

alexandr402 [8]

Answer:

52 mm/s (approximately)

Explanation:

Given:

Initial speed of the projectile is, $u=60.0\ mm/s$

Angle of projection is, $\theta=30.0\°$

Time taken to land on the hill is, $t=4\ s$

In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.

So, the velocity in the horizontal direction always remains the same.

The horizontal component of initial velocity is given as:

$u_x=u\cos\theta\\u_x=60\times \cos(30)\\u_x=30\sqrt3\approx52\ mm/s$

Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.

Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.

6 0

3 years ago

NUUS

Ne4ueva [31]

The answer would be Conduction

4 0

3 years ago

Jill is getting ready to push little Frank in his swing. She pulls Frank back as high as she can and then releases the swing. Wh

makkiz [27]

The transfer of energy is potential energy to kinetic energy. The swing has potential energy when she pulls it back and once she lets go, allowing the swing to move, it then has kinetic energy.

5 0

3 years ago

Two lightbulbs have powers P1 and P2 when separately connected across an ideal battery with potential difference ΔV. The bulbs a

jarptica [38.1K]

Answer:

$P'=\dfrac{P_1P_2}{P_1+P_2}$

Explanation:

Lets take

Resistance of bulb 1 =R₁

Resistance of bulb 2 =R₂

As we know that power P

P= ΔV²/R

Given that voltage difference is same for both bulbs

P₁R₁= ΔV² --------1

P₂R₂= ΔV² -----------2

When these resistance are connected in series then equivalent resistance R

R=R₁+ R₂

The new power P'

P'=ΔV²/R

P'R=ΔV² ------3

From equation 1 ,2 and 3

P'(R₁+ R₂) = ΔV²

$P'\left(\dfrac{\Delta V^2}{P_1}+\dfrac{\Delta V^2}{P_2}\right)=\Delta V^2$

$P'=\dfrac{P_1P_2}{P_1+P_2}$

6 0

3 years ago