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forsale [732]
3 years ago
14

the human nervous system can propagate nerve impulses at about 10 squared metered per second. estimate the time it takes to trav

el 2 m from your toes to your brain.
Physics
2 answers:
Amiraneli [1.4K]3 years ago
8 0

Speed of nervous system

v = 10^2 m/s

distance from toes to head

d = 2 m

now as we know that

speed = \frac{distance}{time}

v = \frac{2}{t}

given that

v = 10^2 m/s

10^2 = \frac{2}{t}

on solving the above equation

t = 0.02 s

<u>so it will take 0.02 seconds to travel</u>

rusak2 [61]3 years ago
7 0
Use the formular d = v x t
d = 2m
v= 100m/s

t= d / v
 = 2 / 100
 = 0.02sec
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What force must be applied to the end of a rod along the x-axis of length 2.25 m in order to produce a torque on the rod about t
4vir4ik [10]

Answer:

<h2>4.9N approx.</h2>

Explanation:

Step one:

given data

length of rod r= 2.25m

Required torque = 11 Nm

Required

The force needed to produce a torque of 11Nm

Step two:

<em>"By definition, Torque is the twisting force that tends to cause rotation. </em>

<em>The point where the object rotates is known as the axis of rotation."</em>

Mathematically,

\tau = rF\sin\theta

\tau = torque

r = radius

F = force

\theta = angle between F and the lever arm

in this case is zero

\tau = rF\\\\F=\tau/r

substituting we have

F=11/2.25

F=4.88N

F=4.9N approx.

7 0
2 years ago
You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
bezimeni [28]

Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to  7.0 k g .m 2  

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

a)

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

<u>ωf = 1.38 rev/sec =</u>

b)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

<u>ΔK.E = 53 J</u>

5 0
3 years ago
Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,
Leviafan [203]

Answer:

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

Explanation:

As we know that the charge per unit length of the long cylinder is given as

\lambda = \frac{q}{L}

here we know that the electric field between two cylinders is given by

E = \frac{2k\lambda}{r}

now we know that electric potential and electric field is related to each other as

\Delta V = - \int E.dr

\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr

\Delta V = -2k \lambda ln(\frac{b}{a})

\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

7 0
3 years ago
A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. Find the magnitude of the
Fittoniya [83]

Answer : 0.814 newton

Explanation:

force (magnetic) acting on the wire is given by

F= ? , I=2.2amp , B = 0.37 T

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4 0
3 years ago
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podryga [215]
Your answer is "<span>surface of a sphere"

Hope this helps.</span>
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