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DiKsa [7]
3 years ago
14

Calculate the wavelength when the frequency of 40,000 Hz travelling at 1450 m/s in fatty tissue.

Physics
1 answer:
dybincka [34]3 years ago
4 0

Answer:

0.036m

Explanation:

Given parameters:

Frequency  = 40000Hz

Speed  = 1450m/s

Unknown:

Wavelength = ?

Solution:

To solve this problem, we use the wave velocity equation:

            V  = F ∧  

V is the velocity

F is the frequency

∧ is the wavelength

  Now insert the parameters and solve;

             1450  = 40000 x ∧

                 ∧ = 0.036m

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A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train th
Rasek [7]

Answer:

0.1575 m/s^2

Explanation:

Solution:-

- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).

- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.

- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).

- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:

                         v_f^2 = v_i^2 + 2*a*( s_f - s_o )

- We will plug in the given parameters in the equation of motion given above:

                         42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}

Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2

7 0
3 years ago
During a normal reaction to a stressful event, muscles are moved to their maximum capacity, and sensitivity is
Aleonysh [2.5K]

Answer:

The paper focuses on the biology of stress and resilience and their biomarkers in humans from the system science perspective. A stressor pushes the physiological system away from its baseline state toward a lower utility state. The physiological system may return toward the original state in one attractor basin but may be shifted to a state in another, lower utility attractor basin. While some physiological changes induced by stressors may benefit health, there is often a chronic wear and tear cost due to implementing changes to enable the return of the system to its baseline state and maintain itself in the high utility baseline attractor basin following repeated perturbations. This cost, also called allostatic load, is the utility reduction associated with both a change in state and with alterations in the attractor basin that affect system responses following future perturbations. This added cost can increase the time course of the return to baseline or the likelihood of moving into a different attractor basin following a perturbation. Opposite to this is the system's resilience which influences its ability to return to the high utility attractor basin following a perturbation by increasing the likelihood and/or speed of returning to the baseline state following a stressor. This review paper is a qualitative systematic review; it covers areas most relevant for moving the stress and resilience field forward from a more quantitative and neuroscientific perspective.

Explanation:

8 0
3 years ago
2.4 What is the radiation error of a temperature measurement?<br>I​
Airida [17]

Answer:

diameter     of       the  wire   =  0.05  in  =0.05 /12   =4.167 *10 ^-3  ft

area   of cross  section    of the  wire  = A =  22/7 *  ( d /2 ) ^2  =0.786 *  (  4.167 *10 ^-3 ) ^2  =1.365 *10 ^-5  ft2  

E  =...

Explanation:

5 0
3 years ago
In an evironmental system of subsystem, the mass balance equation is:__________.
kolbaska11 [484]

Answer:

Explanation:

The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as

Input = Output + Accumulation

The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.

8 0
3 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
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