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2 years ago

Sensory receptors pick up on senses from outside the body. To respond to the sense, what path does the information follow?

2 answers:

8 0

Senses provide information about the body and its environment. Humans have five special senses: olfaction (smell), gustation (taste), equilibrium (balance and body position), vision, and hearing. Additionally, we possess general senses, also called somatosensation, which respond to stimuli like temperature, pain, pressure, and vibration. Vestibular sensation, which is an organism’s sense of spatial orientation and balance, proprioception (position of bones, joints, and muscles), and the sense of limb position that is used to track kinesthesia (limb movement) are part of somatosensation. Although the sensory systems associated with these senses are very different, all share a common function: to convert a stimulus (such as light, or sound, or the position of the body) into an electrical signal in the nervous system. This process is called sensory transduction.

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Radda [10]2 years ago

7 0

Answer:

Follow that smell. SMELL: Originally thought to be present only in the nose, olfactory receptors exist in many locations in the body, including in the skin, heart, lungs, kidneys, muscles, and sperm. ...

Light relaxation. ...

A taste for pathogens

Explanation:

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What sound signal alerts others that a boat is backing up?

inn [45]

"Three short blasts" sound signal alerts others when a boat is backing up based on the boating safety rule. The short blast is a sound which has one second period, thus three short blasts mean three of this kind of sound. The sound signal is used for alerting anything surrounding the boat to avoid a collision.

5 0

3 years ago

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

6 0

3 years ago

Between 1911 and 1990, the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of 1.2 mm/y.T

Anit [1.1K]

Answer:

Angular speed, $\omega=6.90\times 10^{-13}\ rad/s$

Explanation:

It is given that,

The top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of, v = 1.2 mm/yr

$1\ mm/yr=3.171\times 10^{-11}\ m/s$

Velocity, $v=3.80\times 10^{-11}\ m/s$

Height of the tower, h = 55 m

The height of the tower is equivalent to the radius. Let $\omega$ is the angular speed of the tower’s top about its base. The relation between the angular speed and the angular speed is given by :

$v=r\omega$

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{3.80\times 10^{-11}\ m/s}{55\ m}$

$\omega=6.90\times 10^{-13}\ rad/s$

So, the average angular speed of the tower’s top about its base is $6.90\times 10^{-13}\ rad/s$ . Hence, this is the required solution.

6 0

3 years ago

Two identical balls collide one was initially at rest. after the collision one of the balls has a velocity of 3 m/s south. the o

Butoxors [25]

Two identical balls collide head on. The initial velocity of one is 0.75 m/s east, while that of the other one is 0.43 m/s west.

4 0

3 years ago

A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car

Illusion [34]

The horizontal force applied is 160 N while the velocity is 2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

Learn more about force and work:brainly.com/question/758238

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5 0

2 years ago