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A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

nikdorinn [45]

Answer:

a

$\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

b

$I = 0.106 \ A$

Explanation:

From the question we are told that

The current is $I = 0.106 \ A$

The length of one side of the square $a = 4.60 \ cm = 0.046 \ m$

The separation between the plate is $d = 4.0 mm = 0.004 \ m$

Generally electric flux is mathematically represented as

$\phi_E = \frac{Q}{\epsilon_o}$

differentiating both sides with respect to t is

$\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=> $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

Here $\epsilon_o$ is the permitivity of free space with value

$\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=> $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=> $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

Generally the displacement current between the plates in A

$I = 8.85*10^{-12} * 1.1977 *10^{10}$

=> $I = 0.106 \ A$

3 0

3 years ago

In a typical rear-end collision, the victim's ________ is/are accelerated faster and harder than the torso

Georgia [21]

The victim's head is accelerated faster and harder than the torso when the victom is involved in a typical rear-end collision.

The traffic accident where a vehicle crashes into another vehicle that is directly in front of it is called a rear-end collision.

One of the most common accident in the United States is the rear-end collision, and in a lot of cases, rear-end collisions are prompted by drivers who are inattentive, unfavorable conditions of the road, and poor following distance.

An enough room in front of your car so you can stop when the car in front of you stops suddenly is one basic driving rule. The person isn’t driving safely if he / she is behind you and couldn’t stop.

8 0

3 years ago

Read 2 more answers

What is the different between conditions and radiation?

Dmitry_Shevchenko [17]

Answer:

Conduction is the transfer of heat energy by direct process.

Radiation is the transfer of heat energy with electromagnetic waves

Explanation:

5 0

3 years ago

In an insulated container, 0.50 kg of water at 80°C is mixed with 0.050 kg of ice at −5.0°C. After a while, all the ice melts, l

Phoenix [80]

Explanation:

When ice tends to absorb heat energy then it will change into liquid state. Now, change in the temperature of cold water $dT = T_{f} - 0^{o}C$ = $T_{f}$

Change in the temperature of original water in the container $dT' = 80 -T_{f
}$

Latent heat of melting of ice L = 80 cal/g

Specific heat of water C = 1 cal/g C

Heat absorbed by ice to melt = mass( $m_{i}$ ) x latent heat(L)

Heat absorbed by water = mass( $m_{w}$ ) x specific heat(C) x change in temperature (dT')

$m_{i}$ = 0.050 kg = 50 g and $m_{w}$ = 0.50 kg = 500 g

Now, heat lost by water is as follows.

heat lost by hot water( at $80^{o}C$ ) = heat absorbed by the ice + heat absorbed by cold water

$m_{i} \times L + m_{i} \times C dT = m_{w} \times C dT'$

$50 \times 80 + 50 \times 1 \times (T_{f}) = 500 \times 1 \times (80 - T_{f})$

So, we will cancel out 50 from each term given above. Hence,

$80 + T_{f} = 800 - 10T_{f}$

$11 T_{f}$ = 720

$T_{f} = 65.45^{o}C$

Thus, we can conclude that final temperature of the water is $65.45^{o}C$ .

6 0

3 years ago

Read 2 more answers

A cylindrical rod 120 mm long and having a diameter of 15.0 mm is to be deformed using a tensile load of 35,000 N. It must not e

aniked [119]

Answer:

The possible candidates that meet the listed criteria are Titanium Alloy and Steel Alloy as explained further as follows

Explanation:

The given variables are

length of rod = 120 mm

diameter of rod = 15 mm

applied tensile force = 35000 N

limit of diameter reduction = 1.2×10⁻² mm

To asses the listed metals if they meet the two criterion of

1. The metal selected must not experience plastic deformation from the applied load and

2. The reduction in diameter from the applied load should not exceed 1.2×10⁻² mm

For the first criteria we have

Stress = Force/Area where the area = π×r² and r = Diameter/2

We have area = π×0.0075² → and σ = (35000 N)/(π×(0.0075 mm)²) = 198059484.74 Pa or 198.06 MPa

Hence from the listed metals, the ones that meet the first criteria are Aluminum Alloy, Titanium Alloy and Steel Alloy

For the second criterion, we apply Poisson's ratio as follows

v = -εx/ εz = -(Δd/d₀)÷(σ/E), rearranging to make Δd the subject of the formula we have Δd = - (v × σ × d₀)/E

For Aluminum Alloy we have Δd = -(0.33×198.06 MPa×15mm)/(70 GPa) = 1.4×10⁻²mm hence Aluminum Alloy should be excluded

For Titanium Alloy we have Δd = -(0.36×198.06 MPa×15mm)/(105 GPa) = 1.019×10⁻² mm, hence Titanium alloy meets the requirements

For the Steel Alloy we have Δd = -(0.27×198.06 MPa×15mm)/(205 GPa) = 3.9×10⁻³mm Hence the steel alloy also meets the requirement

5 0

3 years ago