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3 years ago

Which of these receptor types functions as an exteroceptor?

Physics

2 answers:

horrorfan [7]3 years ago

5 0

Answer:

All of the listed responses are correct.

Explanation:

Stimuli in the environment activate specialized receptor cells in the peripheral nervous system. Different types of stimuli are sensed by various types of receptor cells. Receptor cells can be classified into types on three different criteria:

1.cell type

2.position,

3. function.

Receptors can be classified structurally on the basis of cell type and their position in relation to stimuli they sense. classification can be functionally on the basis of the transduction of stimuli, or how the mechanical stimulus, photo, or chemical changed the cell membrane potential. An exteroceptor is a receptor that is located near a stimulus in the surrounding environment, example is the somatosensory receptors that can be found in the dermis

AnnZ [28]3 years ago

5 0

Answer:

Explanation:

All of the structures illustrated are sensitive to stimuli arising outside the body

An exteroceptor is a receptor that is located near a stimulus in the external environment, such as the somatosensory receptors that are located in the skin.

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If a refrigerator is heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the r

aliya0001 [1]

The first law of thermodynamics is a version of the law of conservation of energy, adapted for thermodynamic systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed. so assuming no heat losses then heat removed is also 333 J

5 0

3 years ago

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0

3 years ago

Law of conservation of energy states that

SCORPION-xisa [38]

A) energy cannot be created nor destroyed

4 0

3 years ago

Read 2 more answers

If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Arada [10]

The force on the fry is $0.015 N$

Explanation:

We can find the force acting on the fry by using Newton's second law:

$F=ma$

where

F is the net force on the fry

m is its mass

a is its acceleration

For the fry in this problem,

$m=3 g = 0.003 kg$

$a=5.0 m/s^2$

Therefore, the force exerted on the fry is

$F=(0.003)(5)=0.015 N$

Learn more about Newton's second law here:

brainly.com/question/3820012

#LearnwithBrainly

3 0

3 years ago

A lump of clay whose rest mass is 4 kg is travelling at three-fifths the speed of light when it collides head-on with an identic

Harman [31]

Answer:

mass of the composite lump is 10 kg

Explanation:

given data

mass = 4 kg

to find out

mass of composite lump

solution

we know energy is conserved so

so m1 = m2 = m0 that is 4kg

and

E(1) release+ E(2) release = E(1,2) rest

so γ(1)m(1)c² + γ(2)m(2)c² = Mc² ..........................1

that why here

|v(1)| = |v(2)| = 3/5 c ......................2

and

γ = 1 / √(1 − v²/c²) .......................3

put here v = 3 and c is 5

γ = 1 /√(1 − 9/25)

γ = 5/4

γ(1) = γ(2) = γ = 5/4

so from equation 1

γ(1)m(1)c² + γ(2)m(2)c² = Mc²

M = 2γm0

M = 2(5/4 )(4)

M = 10 kg

so mass of the composite lump is 10 kg

7 0

4 years ago