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Mumz [18]
3 years ago
14

Consider two thin disks, of negligible thickness, of radiusR oriented perpendicular to thex axis such that the x axis runs throu

gh thecenter of each disk. The disk centered at x=0 has positive charge density eta, and the disk centered at x=a has negative charge density -\eta, where the charge density is charge perunit area. What is the magnitude E of the electric field at the point on thex axis with x coordinate a/2? For what value of the ratio R/a of plate radius to separation between the plates does the electric field at the point x=a/2 on the x axis differ by 1 percent from the result η/ϵ for infinite sheets?

Physics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Complete question

The complete question is shown on the first and second uploaded image

Answer:∈

Answer to first question is shown on the second uploaded image.

Part B the Answer is:

The ratio  \frac{R}{a} is evaluated to be 49.99

Explanation:

The explanation is shown on the third ,fourth and fifth image.

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An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the
hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg

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3 years ago
When children use the word “doggie” to refer to every four-legged animal they see, __________ occurs.
Oduvanchick [21]
This is known as overextension! (the correct answer is B.)
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4 years ago
A 230 kg steel crate is being pushed along a cement floor. The force of friction is 480 N to the left and the applied force is 1
tamaranim1 [39]
The acceleration would be 6m/sThis is because of the formula, "f/m=a" to find the acceleration; We would need to subtract the force of the friction which equals 1380, then divide that by the mass (which was 230) to get the answer 6
7 0
3 years ago
Read 2 more answers
A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t
WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s

So, the required time is 2.2 seconds.

3 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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