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ArbitrLikvidat [17]
3 years ago
9

Ultrasound with a frequency of 4.077 mhz can be used to produce images of the human body. if the speed of sound in the body is t

he same (2.31 km/s) as in salt water, what is the wavelength in the body? answer in units of m. 01
Physics
1 answer:
mestny [16]3 years ago
3 0
The frequency of the ultrasound is (converting in Hz)
f=4.077 MHz=4.077 \cdot 10^6 Hz
while the speed of sound in the body is
v=2.31 km/s = 2310 m/s

We can calculate the wavelength of the ultrasound in the body by using the relationship betwen frequency, wavelength and speed of the wave:
\lambda =  \frac{v}{f}= \frac{2310 m/s}{4.077 \cdot 10^6 Hz}=5.67 \cdot 10^{-4} m
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Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

4 0
3 years ago
You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use
Yuri [45]
P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
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3 0
3 years ago
A source charge generates an electric field of 4286 N/C at a distance of 2. 5 m. What is the magnitude of the source charge? (Us
svp [43]

The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.

What does Gauss Law state?

It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.

Q = \dfrac {ER^2}k

Where,

E = electric force = 4286 N/C

k = Coulomb constant = 8.99 \times  10^9 \rm\ N m ^2 /C ^2

Q\\
     = charges = ?

r = distance of separation = 2.5 m

Put the values in the formula,

Q  = \dfrac {4286\times  2.5 ^2}{8.99 \times  10^9 }\\\\
Q  = 3\rm \  \mu C

Therefore, the magnitude of the source charge is 3 μC.

Learn more about Gauss's law:

brainly.com/question/1249602

8 0
2 years ago
A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of
damaskus [11]
The parallel component is given by
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3 years ago
The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t
balu736 [363]

Answer:

D=2.996\times 10^{-2} m

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as  1.0\times 10^5 N/C.  transferring 3.9\times 10^9 electrons, the disk's Area can be calculated using the formula:

E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2

#We now calculate the disks diameter:

A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m

Hence, the diameter of the disks is D=2.996\times 10^{-2} m

8 0
3 years ago
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