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2 years ago

6

The sensory portion of the pns carries electrical signals __ the cns; the motor portion carries electrical signals _

_ the cns.

1 answer:

tatuchka [14]2 years ago

8 0

I think the correct word to fill in the blanks is to and from, respectively. The sensory portion of the pns carries electrical signals to the cns; the motor portion carries electrical signals from the cns. Hope this answers the question. Have a nice day.

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What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Stolb23 [73]

Answer:

$1.1\cdot 10^{-10}C$

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

$q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C$

8 0

2 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

A positively-charged piece of plastic exerts an attractive force on an electrically neutral piece of paper. This is because: 1.

Furkat [3]

Answer:

1 ) Electrons are less massive than than atomic nuclei.

Explanation:

A positively charged body tends to attract negatively charged particle and repel positively charged particle. Neutral body consists of atoms which contain both positively charged particles ( electrons ) and negatively charged particles ( protons ). Electrons are small and light in weight . Both electrons and protons experience equal and opposite force by an external charged body but shift in electron is more because of their being comparatively lighter. So the body gets polarized due to uneven distribution of charge. This results into body getting attracted through the process of induction.

3 0

2 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

4 0

2 years ago

A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar

Debora [2.8K]

Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

$0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700$

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

$d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m$

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

8 0

2 years ago