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Vladimir79 [104]
3 years ago
8

A 25.0-meter length of platinum wire with a cross-sectional area of 3.50 × 10^−6 meter^2 has a resistance

Physics
1 answer:
Nookie1986 [14]3 years ago
4 0
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m
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Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,
Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m

The frequency of a wave is given by

f=\frac{v}{\lambda}

where

v is the speed of the wave

\lambda is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

\lambda=0.004 m is the wavelength

So, the frequency is

f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

f'=\frac{v\pm v_r}{v\pm v_s}f

where

f' is the apparent frequency

v is the speed of the wave

v_r is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

v_s is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

v_r = +v (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

v_s = -v

also

v = c = speed of sound in the blood

So the formula becomes

f'=\frac{c + v}{v - v_s}f

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, \rho.

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

3 0
3 years ago
Types of mechanical waves include
USPshnik [31]
I think surface waves
4 0
3 years ago
3. A 6 kg block moving to the right at 4 m/s collides with and sticks to a stationary block of unknown mass. If the two blocks m
Sphinxa [80]

Answer: 2kg

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This problem is a textbook conservation of momentum problem. The intial momentum is equal to the final momentum. For the initial state of each block, only the first one was moving. Then they both combine to move together.

Pi = Pf

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6 0
3 years ago
PLEASE HLEP I WILL GIVE BRAINLIEST TO CORRECT ANSWER!!!!!!
Hoochie [10]

Answer:

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8 0
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A ceramic tile measuring 50 cm x50cm has been designed to bear a pressure of 40 N/in . Will it with stand a force of 5 N?
dangina [55]

Answer:

The ceramic tile will stand the force of 5 N.

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Step 1: Calculate the area (A) of the ceramic tile

We will use the formula for the area of a square.

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Step 2: Convert "A" to in²

We will use the conversion factor 1 in² = 6.45 cm².

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Step 3: Calculate the pressure (P) exerted by a force (F) of 5 N

We will use the following expression.

P = F/A

P = 5 N / 3.9 × 10² in² = 0.013 N/in²

Since the pressure exerted would be less than the maximum pressure resisted (40 N/in²), the ceramic tile will stand the force of 5 N.

3 0
3 years ago
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