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IRINA_888 [86]
4 years ago
6

Brain surgery. Surgeons can remove brain tumors by using a cavitron ultrasonic surgical aspirator, which produces sound waves of

frequency 23.0 kHzkHz . What is the wavelength of these waves in air
Physics
1 answer:
dmitriy555 [2]4 years ago
7 0

Answer:\lambda =1.304\times 10^4\ m

Explanation:

Given

frequency of wave \nu =23\ kHz

We know velocity is given by

v=\nu \times \lambda

where \lambda=wavelength

\lambda =\frac{velocity}{\nu }

\lambda =\frac{3\times 10^8}{23\times 10^3}

\lambda =1.304\times 10^4\ m

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infrared radiation

Explanation:

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3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
Goshia [24]

Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

6 0
3 years ago
point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is
Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, \rho_{w} = 1000\ kg//m^{3}

Acceleration due to gravity, g = 9.8\ m/s^{2}

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = \frac{H + D}{2}

h = \frac{3 + 2.5}{2} = 2.75\ m

Volume of water in the pool, V = \pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}

Mass of water, m_{w} = \frac{\rho_{w}}{V}

m_{w} = \frac{1000}{311.02} = 3.215\ kg

Work done is given by the potential energy of the water as:

W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done,W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J

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3 years ago
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