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babymother [125]

3 years ago

7

A-delta fibers : A) are small, myelinated fibers. B) transmit pain signals at a slower rate than C-fibers. C) typically transmit

dull, aching pain. D) do not respond to mechanical pai

1 answer:

daser333 [38]3 years ago

7 0

Answer:

A

Explanation:

Aδ fibers carry cold, pressure, and acute pain signals, and because they are thin (2 to 5 μm in diameter) and myelinated, they send impulses faster than unmyelinated C fibers, but more slowly than other, more thickly myelinated group A nerve fibers. Their conduction velocities are moderate.

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to

Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

$p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa$

On the lower part of the hatch, there is a pressure equal to

$p_{bot}=p_{atm}=1.013\cdot 10^5 Pa$

So, the net pressure acting on the hatch is

$p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa$

which acts from above.

The area of the hatch is given by:

$A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2$

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

$F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N$

8 0

3 years ago

what is the name of the area around a charged object where the object can exert a force on other charged objects?

shusha [124]

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

4 0

2 years ago

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart

natita [175]

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,

Force of gravity = magnetic force

$mg=ilB$ ..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$ , v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0

3 years ago

A concert loudspeaker suspended high off the ground emits 31 W of sound power. A small microphone with a 1.0 cm^2 area is 42 m f

Aleonysh [2.5K]

Answer:

intensity of sound at level of microphone is 0.00139 W / m 2

sound intensity level at position of micro phone is 91.456 dB

EXPLANATION:

Given data:

power of sound P = 31 W

distance betwen microphone & speaker is 42 m

a) intensity of sound at microphone is calculated as

$I = \frac{P}{A}$

$= \frac{34}{4 \pi ( 44m )^ 2}$

= 0.00139 W / m 2

b) sound intensity level at position of micro phone is

$\beta = 10 log \frac{I}{I_o}$

where I_o id reference sound intensity and taken as

$= 1 * 10^{-12} W / m 2$

$\beta = 10 log\frac{0.00139}{10^[-12}}$

= 91.456 dB

7 0

3 years ago

A 480 g peregrine falcon reaches a speed of 69 m/s in a vertical dive called a stoop. If we assume that the falcon speeds up und

LiRa [457]

Answer:

The minimun height is 242 [m]

Explanation:

We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy. We will take the point where the Falcon reaches the speed of 69 (m/s), as the point where the potential energy is zero, i.e. it will be the reference point.

At the reference point all potential energy has been transformed into kinetic energy, therefore the kinetic energy can be calculated.

$E_{k}=0.5*m*v^{2} \\ where:\\v = velocity = 69 [m/s]\\m = mass = 480[g] = 0.480[kg]\\E_{k} = kinetic energy [J]\\E_{k} =0.5*0.48*(69)^{2} \\E_{k} =1142.64[J]$

Now we can calculate the elevation with respect to the reference point using the definition of the potential energy.

$E_{p}=m*g*h\\ E_{p}=E_{k} \\therefore\\h= E_{p}/(m*g)\\h= 1142.64/(.48*9.81)\\h=242[m]$

4 0

3 years ago