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2 years ago

15

At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc

reasing at a rate of 3.0 J/s, how fast is the current changing

1 answer:

lys-0071 [83]2 years ago

5 0

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

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Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

8 0

2 years ago

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) <u>On doubling the mass:</u>

New mass, $m'=2m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

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1 year ago

A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro

Lerok [7]

Answer:

0.0000045 s

Explanation:

f = Frequency = 8 MHz

Clock cycle is given by

$\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s$

Time taken for 12 clock cycles

$12\times 1.25\times 10^{-7}=0.0000015\ s$

Time taken per instruction is 0.0000015 s

In reading and displaying information it requires 3 processes

1 for reading, 1 for searching and 1 for displaying.

$3\times 0.0000015=0.0000045\ s$

Time taken is 0.0000045 s

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2 years ago

Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.

Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

$Q=-P_{tot}=-(P_{2}-P_{1})$

The potential energy for any point can be calculated as:

$P=m*g*h$

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

$Q=P_{1}=m*g*h$

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

$W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW$

5 0

2 years ago

A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material

worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

$r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m$ is the radius

$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

$d=\frac{m}{V}$

where m = 1 kg is the mass. Substituting, we find

$d=\frac{1 kg}{7.1\cdot 10^{-5} m^3}=14,160 kg/m^3$

3 0

2 years ago