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belka [17]
3 years ago
14

You are viewing a read out of an oscilloscope that shows you when different voltages in a circuit maximize. If the period of the

circuit is 0.04 seconds, how much time is there between when the capacitor voltage maximizes and the inductor voltage maximizes ?
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
7 0

Answer:

The time is  t = 0.02 \ s  

Explanation:

From the question we are told that

  The period of the circuit is  T  =  0.04

Generally voltage maximization of the capacitor occurs during the voltage minimization of the inductor and vise versa

So the time between the voltage maximization of the capacitor and that of the inductor is mathematically represented as

           t = \frac{T}{2}

=>        t = \frac{0.04}{2}

=>        t = 0.02 \ s  

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What did the asymptote say to the removable discontinuity worksheet answers?
ra1l [238]

“Don't hand that holier than thou line to me” is what the asymptote said to the removable discontinuity.

 

 

The distance between the curve and the line where it approaches zero as they tend to infinity is the line in the asymptote of a curve. This is unusual for modern authors but in some sources the requirement that the curve may not cross the line infinitely often is included.

 

The point that does not fit the rest of the graph or is undefined is called a removable discontinuity. By filling in a single point, the removable discontinuity can be made connected.

6 0
3 years ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
8090 [49]

Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

Explanation:

Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

Electron mass = 9.11×10^-31 Kg

(a) What is the speed of the electron 1.3 ns after entering this region?

E = F/q

F = Eq

Ma = Eq

M × V/t = Eq

Substitute all the parameters into the formula

9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19

V = 7.68×10^-18 /7.0×10^-22

V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

3 0
3 years ago
A box sits on a table. A long arrow labeled F subscript P = 22 N points right. A short arrow labeled F subscript f = ? N points
anygoal [31]

Answer:

b -4 N

may the force be with you

8 0
3 years ago
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