A mechanoreceptor in the papillary layer of the dermis that responds to fine touch is a

Physics

1 answer:

yan [13]4 years ago

5 0

Answer: tactile (Meissener) corpuscle.

Hope this helps.

You might be interested in

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

5 0

4 years ago

if the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity?

erastova [34]

The answer to this question should be: The accuracy in measuring its velocity decreases

Hope I helped

7 0

3 years ago

1. which of the following should be classified as a suspension?

Mumz [18]

Where are the answers to this question?

3 0

4 years ago

If the mass of a

mixer [17]

Answer:

The final acceleration becomes (1/3) of the initial acceleration.

Explanation:

The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

$F=ma$

m = mass

a = acceleration

According to given condition, if the mass of a sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

$a=\dfrac{F}{m}$