3 years ago

Which of the following epidermal cells are the sensory receptors for touch?

Physics

1 answer:

never [62]3 years ago

7 0

Merkel cells are the sensory receptors for touch.

You might be interested in

What conditions can create the largest waves in the ocean. this is for science. i need done asap.

Yuki888 [10]

The answer is strong winds, i hoped this helped.

→if this helped please mark brainliest i need to level up←

7 0

2 years ago

While traveling along a highway a driver slows from 31 m/s to 15 m/s in 8 seconds. What is the automobile’s acceleration? (Remem

MaRussiya [10]

Answer:

The automobile's acceleration in that time interval is -2 m/s^2

Explanation:

The acceleration is defined as the rate of change of the velocity.

The average acceleration in a given lapse of time is calculated as:

A = (final velocity - initial velocity)/time.

In this case, we have:

initial velocity = 31 m/s

final velocity = 15 m/s

time = 8 seconds.

Then the average acceleration is:

A = (15m/s - 31m/s)/8s = -2 m/s^2

8 0

2 years ago

A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a

kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

$\lambda = 720 \ nm$

or,

$= 720\times 10^{-9} \ m$

$D=0.85 \ m$

$d = 0.22 \ mm$

or,

$=0.22 \times 10^{-3} \ m$

As we know,

Fringe width is:

⇒ $\beta=\frac{\lambda D}{d}$

hence,

Separation between second and third bright fringes will be:

⇒ $\theta=\beta=\frac{\lambda D}{d}$

$=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}$

$=2.78\times 10^{-3} \ m$

or,

$=2.78 \ mm$

8 0

2 years ago

how much chemical energy must be supplied to a car engine if we want it to produce 5000J of kinetic energy and it is 45% efficie

vagabundo [1.1K]

Here i state the conservation of energy rule and use that to justify my answer. I showed how to manipulate percentages to get the final answer of 11000J (2sf). Hope I'm right xx

3 0

3 years ago

A light ray transfers from more to less dense medium at certain condition , if we repeat this experiment by increasing the angle

evablogger [386]

Answer:

The correct option is;

Still constant

Explanation:

The relative refractive index ₁n₂ between the two medium can be as follows;

$_1n_2 = \dfrac{The \ speed \ of \ light \ in \ medium \ 1}{The \ speed \ of \ light \ in \ medium \ 2} = \dfrac{c_1}{c_2}$

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.

4 0

2 years ago