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3 years ago

A neuron that is activated when a mosquito lands on your arm

1 answer:

8 0

<span> The spinal neurons involved in the tingling sensation caused by a light touch are different from those transmitting pain or a 'chemical' itch, the latter elicited by a mosquito bite or a skin wound that is healing.
hope this helps</span>

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A cyclical motion occurs because of density differences in the mantle. Heated, less dense lower regions of the fluid mantle rise

Marta_Voda [28]

The answer for this is B

3 0

3 years ago

Which of the following determines and objects ability to float in water

AleksandrR [38]

I think its
Mass and volume

6 0

3 years ago

Particles of m1 and m2 (m2>m1) are connected by a line in extensible string passing over a smooth fixed pulley. Initially, bo

Tresset [83]

Answer:

The velocity with which the mass will hit the floor is $v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.}$

Explanation:

If the tension in the string is $T$ , for $m_1$ we have

$T- m_1g =m_1a$ ,

and for the mass $m_2$

$T -m_2g = -m_2a$

From these equations we solve for $a$ and get:

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

The kinematic equation

$v_f^2 = v_0^2+2ax$

gives the final velocity $v_f$ of a particle, when its initial velocity was $v_0$ , and has traveled a distance $x$ while undergoing acceleration $a$ .

In our case

$v_0 = 0$ (the initial velocity of the particles is zero)

$a =(\dfrac{m_2-m_1}{m_2+m_1}) g.$

which gives us

$v_f^2 = 2ax$

$v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g$

$\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }$

which is the velocity with which the mass $m_2$ will hit the floor.

8 0

3 years ago

A 65.0 kg skier is moving at 6.85 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 4.00

marusya05 [52]

Answer:

v = 4.58 m/s

Explanation:

In order to calculate the speed of the skier when she gets the bottom of the hill, you have to calculate the speed of the skier when she crosses the rough patch.

To calculate the velocity at the final of the rough patch you take into account that the work done by the friction surface is equal to the change in the kinetic energy of the skier:

$W_f=\Delta K\\\\-N\mu_kd=\frac{1}{2}mv^2-\frac{1}{2}mv_o^2=\frac{1}{2}m(v^2-v_o^2)$ (1)

Where the minus sign means that the work is against the motion of the skier.

Wf: friction force

m: mass of the skier = 65.0kg

N: normal force = mg

g: gravitational acceleration = 9.8m/s^2

d: distance of the rough patch = 4.00m

v: speed at the end of the rough patch = ?

vo: initial speed of the skier = 6.85m/s

μk: coefficient of kinetic friction = 0.330

You replace the expression for the normal force in the equation (1), and solve for v:

$-mg\mu_kd=\frac{1}{2}m(v^2-v_o^2)\\\\-g\mu_kd=\frac{1}{2}(v^2-v_o^2)\\\\v=\sqrt{-2g\mu_kd+v_o^2}\\\\v=\sqrt{-2(9.8m/s^2)(0.330)(4.00m)+(6.85m/s)^2}=8.53\frac{m}{s}=4.58\frac{m}{s}$

Then, the speed fot he skier at the bottom of the hill is 4.58m/s

3 0

4 years ago

How much force is needed to stop a 80 kg football player if he decelerates at 5m/s?

cricket20 [7]

Use Newton's 2nd law of motion.
Here it is:

   Force = (mass) x (acceleration)

   = (80 kg) x (-5 m/s²)

   =    -400 Newtons .

I called the acceleration negative because the player is slowing down.

The force comes out negative because it's in the direction opposite to
his motion.

4 0

4 years ago