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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
Physics
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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<span>Cell phone services can be compared through several characteristic. The rate of the data transfer: the higher the rate the better the service, the availability- the percent of the time the service is available (the greater the percent the better the service) and the reliability, which means that the service is non interrupted.</span>



4 0
3 years ago
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

6 0
4 years ago
An audi travels at 88 km/hr over a distance of 22 km. calculate.25 the time taken for the audi to travel this distance.
sammy [17]
The Audi covers 88 in one hour.
22 is one quarter of 88, so it takes one quarter of one hour. That's 15 minutes.
Note:. The answer would be the same for any other brand of car, motorcycle, airplane etc.
4 0
4 years ago
A basketball player is getting ready to jump, pushing off the ground and accelerating upward.
lidiya [134]

Answer:

B

Explanation:

B

3 0
3 years ago
A 6 kg bowling ball is accelerated at a rate of 2.3 m/s² down the lane. How much force was necessary to produce this acceleratio
Maurinko [17]
The answer is D hoped this helped
4 0
3 years ago
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