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snow_tiger [21]
3 years ago
8

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

are reflected from a metal fragment in the muscle and are detected by the transducer 1.5 x 10−4 seconds after they were first emitted. What is the depth of the metal fragment in the muscle tissue?
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

0.195 m

Explanation:

Speed is distance moved per unit time, expressed as s=d/t and making d the subject of the formula then d=st

Where d is distance/depth moved, s is rhe speed of waves and t is time in seconds.

Substituting s with 1300 m/s and t with 0.00015 s then the depth of metal segment will be

D=1300*0.00015=0.195 m

Therefore, the depth is equivalent to 0.195 m

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The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi
lukranit [14]

Answer:

a. volume of gas:  (decreases)

b. temperature of gas:  (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

4 0
3 years ago
A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically eq
andreyandreev [35.5K]

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>

time mass passes through equilibrium = 1/12 seconds = 0.083

<u>b) Calculate the time at which the mass attains its extreme displacement </u>

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

<u>c) What is the position of the mass at this instant</u>

position = 3e^-4/3

attached below is the detailed solution to the given problem

6 0
3 years ago
Air rushing against an airplane is an example of _____friction.
slava [35]
The word "static" would be known to be friction as air rushing against an airplane
6 0
3 years ago
what output force is generated when an input force of 630 n is applied to a machine with a mechanical advantage of 3
Marta_Voda [28]
The mechanical advantage is the factor by which
the machine multiplies the input force.

If the MA is 3 and the input force is 630N, then
the output force is

           (3) x (630N) = 1,890N
3 0
3 years ago
Read 2 more answers
The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s
just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c)  78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0
3 years ago
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