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1 year ago

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The oscillating current in an electrical circuit is as follows, where I is measured in amperes and t is measured in seconds. I =

7 sin(60πt) + cos(120πt). Find the average current for each time interval. (Round your answers to three decimal places.) (a) 0 ≤ t ≤ 1 60 amps

1 answer:

Vesna [10]1 year ago

3 0

Answer:

Explanation:

Given

$I=7\sin (60\pi t)+\cos (120\pi t)$

$=\frac{1}{\frac{1}{60}}\times \int_{0}^{\frac{1}{60}}7\sin (60\pi t)dt+\int_{0}^{\frac{1}{60}}\cos (120\pi t)dt$

$=60\left [ \frac{-7}{60\pi }\cos (60\pi t)+\frac{1}{120\pi }\sin (120\pi t)\right ]_{0}^{\frac{1}{60}}$

$=60\left [ \frac{-7}{60\pi }\left ( \cos \pi-\cos 0\right )+\frac{1}{120\pi }\left ( \sin (2\pi )-\sin (0)\right )\right ]$

$=\frac{-7}{\pi }\left ( -1-1\right )+\frac{1}{2\pi }\left ( 0-0\right )$

$=\frac{14}{\pi }$

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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 $b=1$

7 0

1 year ago

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this co

natima [27]

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.

Explanation:

Measure unstretched length of spring, L. E.g. L = 0.60m.

Set mass to a convenient value (e.g. m = 0.5kg).

Hang mass.

Measure new spring length, L'. E.g. L' = 0.70m.

Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m

Use mg = ke (in equilibrium weight = tension)

k = mg/e

Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).

k = 0.5*10/0.10 = 50 N/m

Repeat for a few different masses. (L always stays the same.)

Take the average of your k values.

5 0

1 year ago

Read 2 more answers

An electric clock is hanging on a wall. As you are watching the second hand rotate, the clock's battery stops functioning, and t

Setler [38]

Answer:

B. W is positive and a is negative

Explanation:

As we know that the angular speed of the second clock is in positive direction so as it comes to halt from its initial direction of motion then we have

initial angular velocity is termed as positive angular velocity

$\omega = positive$

now it comes to stop so angular acceleration is taken in opposite to the direction of angular speed

so we will have

$\alpha = negative$

so here correct answer is

B. W is positive and a is negative

8 0

1 year ago

a 59kg physics student jumps off the back of her laser sailboat (42kg). after she jumps the laser is found to be travelling at 1

evablogger [386]

From the conservation of linear momentum of closed system,

Initial momentum = final momentum

Mass of the student, M = 59 kg

Mass of the laser boat, m = 42 kg

Initial speed of student + laser boat, u =0

Final speed of laser boat, v = 1.5 m/s

Final speed of the student = V

(M+m) u =M V +m v

0 = (59 kg) V + (42 kg) (1.5m/s)

V = - 1.06 m/s

Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.

5 0

1 year ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$

We know that $K_c$ is 33Mpa*m^{0.5} and our Safety factor is 2,

$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

$F_{allow}=\sigma_{allow}*L_c*W$

$F_{allow} = 208.15*250*4$

$F_{allow} = 208.15kN$

That is the maximum tensile load before a catastrophic failure.

4 0

1 year ago