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2 years ago

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A 2. 7-cm-diameter parallel-plate capacitor has a 1. 8 mm spacing. The electric field strength inside the capacitor is 1. 2×105

v/m.

1 answer:

Luda [366]2 years ago

7 0

Hi how are u that is included

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Help with either I don’t understand this I know W=f*d but don’t get how it applies here

rosijanka [135]

Answer:

5) Displacement = +3.125 m

Displacement is in the same direction as the force vector.

6) Force = -53.89 N

Force is in an opposite direction relative to the displacement.

Explanation:

5) We are given;

Force; F = 160 N.

Workdone; W = +500 J

Now, formula for workdone is;

W = Force × displacement

Thus, displacement = Work/force

Displacement = 500/160

Displacement = +3.125 m

Thus, displacement is in the same direction as the force vector.

6) We are given;

Displacement; d = 18 m.

Workdone; W = -970 J

Like in the first answer above,

Workdone = Force × Displacement

Thus;

Force = Workdone/Displacement

Force = -970/18

Force = -53.89 N

Since force is negative and displacement is positive, it means force is in an opposite direction relative to the displacement.

3 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0

3 years ago

The higher the temperature of an object the

Allushta [10]

higher temp = higher energy = higher frequency = shorter wavelength

4 0

3 years ago

Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,

Paladinen [302]

Answer:

$0.05997\ m^3/s$

$3596.1395\ L/min$

$2.11647\ ft^3/s$

Explanation:

$1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s$

Volume rate of flow is $0.05997\ m^3/s$

$1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min$

Volume rate of flow is $3596.1395\ L/min$

$1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s$

Volume rate of flow is $2.11647\ ft^3/s$

6 0

3 years ago

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current

balu736 [363]

Answer:

(a) The resistance R of the inductor is 2480.62 Ω

(b) The inductance L of the inductor is 1.67 H

Explanation:

Given;

emf of the battery, V = 16.0 V

current at 0.940 ms = 4.86 mA

after a long time, the current becomes 6.45 mA = maximum current

Part (a) The resistance R of the inductor

$R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms$

Part (b) the inductance L of the inductor

$\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}$

where;

L is the inductance

R is the resistance of the inductor

t is time

$L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H$

Therefore, the inductance is 1.67 H

5 0

3 years ago