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erastova [34]
1 year ago
11

Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a

re given equal but opposite uniform surface charge densities . Ignore edge effects and use Gauss's law to find for points far from the edges, the electric field between the plates. Find the field outside the plates on either side.
Physics
1 answer:
mote1985 [20]1 year ago
4 0

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

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A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
20 points please help!!
IgorC [24]

Answer:

Sample Response: If temperature and surface area increase, then the time it takes for sodium bicarbonate to completely dissolve will decrease, because increasing both factors increases the rate of a chemical reaction.

Explanation:

4 0
2 years ago
The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o
MrMuchimi

Answer:

The total kinetic energy of both particles is 2.43\times10^{-13}

Explanation:

Given that,

Kinetic energy of nucleusK.E= 1.24\times10^{-13}\ J

Kinetic energy of proton K.E= 2.47\times10^{-13}\ J

Radius of proton r= 0.9\times10^{-15}\ m

We need to calculate the final potential energy

Using formula of final potential energy

U=\dfrac{kq^2}{r}

Put the value into the formula

U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}

U_{f}=1.28\times10^{-13}\ J

We need to calculate the initial energy of both the particles

Using formula of energy

E_{i}=(K.E_{n}+K.E_{p})+U_{i}

E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0

E_{i}=3.71\times10^{-13}\ J

We need to calculate the total kinetic energy of both particles

Using conservation of energy

E_{i}=E_{f}

E_{i}=K.E_{f}+U_{f}

3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}

K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}

K.E_{f}=2.43\times10^{-13}

Hence, The total kinetic energy of both particles is 2.43\times10^{-13}

3 0
1 year ago
Starting at point 0, you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees. What is your heigh
ira [324]

Answer:

43.58 m

Explanation:

If you travel 500 m on a straight road that slopes upward at a constant angle of 5 degrees

Using trigonometry ratio

Sin 5 = opposite/hypothenus

Where the hypothenus = 500m

Opposite = height h

Sin 5 = h/500

Cross multiply

500 × sin 5 = h

h = 500 × 0.08715

h = 43.58m

Therefore, the height above the starting point is equal to 43.58m

5 0
2 years ago
The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em
zepelin [54]

Answer:    the answer is d

Explanation: there are not more than 10 violations  within a twelve month period hope this helps

4 0
2 years ago
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