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slavikrds [6]
3 years ago
12

What is the difference between a sonogram and a frequency spectrum

Physics
1 answer:
bagirrra123 [75]3 years ago
5 0
Sonogram is a device that transmit high frequencies to project certain images (ultrasound) , while on the other hand, frequency spectrum  uses certain frequencies to transmit data or information ( radio, cellphone).

Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
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A lava flow is an example of what igneous rock
nydimaria [60]
The answer is extrusive. 
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4 years ago
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A 480 g peregrine falcon reaches a speed of 69 m/s in a vertical dive called a stoop. If we assume that the falcon speeds up und
LiRa [457]

Answer:

The minimun height is 242 [m]

Explanation:

We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy. We will take the point where the Falcon reaches the speed of 69 (m/s), as the point where the potential energy is zero, i.e. it will be the reference point.

At the reference point all potential energy has been transformed into kinetic energy, therefore the kinetic energy can be calculated.

E_{k}=0.5*m*v^{2} \\ where:\\v = velocity = 69 [m/s]\\m = mass = 480[g] = 0.480[kg]\\E_{k} = kinetic energy [J]\\E_{k} =0.5*0.48*(69)^{2} \\E_{k} =1142.64[J]

Now we can calculate the elevation with respect to the reference point using the definition of the potential energy.

E_{p}=m*g*h\\ E_{p}=E_{k} \\therefore\\h= E_{p}/(m*g)\\h= 1142.64/(.48*9.81)\\h=242[m]

4 0
3 years ago
In this unit, the amount of solute is measured in?
Hoochie [10]
Try liters if you haven’t done it yet. I’m so sorry if i’m incorrect.
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3 years ago
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Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
Fill in the blanks HELP ASAP
Alexandra [31]

checking to see if im locked out

7 0
3 years ago
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