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4 years ago

6

1. When is positive work done on a system? When is negative work done?

1 answer:

FinnZ [79.3K]4 years ago

3 0

Answer:

Both positive work and negative work have meaning: Positive work follows when the force has a component parallel to the displacement. Positive work adds energy to a system. Negative work follows when the force has a component opposite or against the displacement.

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A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz.

Pani-rosa [81]

Answer:

$X_c=-33.86275385\Omega$

$|Z|=105.5778675\Omega$

$I=0.04735841062A$

$\phi=20.78612878\°$

Explanation:

The electrical reactance is defined as:

$X_c=-\frac{1}{2\pi fC}$

Where:

$f=Frequency\\C=Capacitance$

So, replacing the data provided by the problem:

$X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega$

Now, the impedance can be calculated as:

$Z=R+jX_c$

Where:

$R=Resistance\\X_c= Capacitive\hspace{3}reactance$

Replacing the data:

$Z=100-j33.86275385$

In order to find the magnitude of the impedance we can use the next equation:

$|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega$

We can use Ohm's law to find the current:

$V=I*Z\\I=\frac{V}{Z}$

Therefore the current is:

$I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j$

And its magnitude is:

$|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega$

Finally the phase angle of the current is given by:

$\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°$

5 0

3 years ago

A weightlifter completes a series of lifts with a 700 N weight. In one lift, he raises the weight to a height of 2.5 m off the g

zhannawk [14.2K]

Answer:

The output power the weightlifter is 2916.67 W.

Explanation:

Given;

weight lifted, W = 700 N

height the weight is lifted, h = 2.5 m

time taken to lift the weight, t = 0.60 s

The output power the weightlifter is calculated as;

Power = Energy applied / time taken

Energy applied = weight lifted x height the weight is lifted

Energy applied = 700 x 2.5

Energy applied = 1750 J

Power = 1750 / 0.6

Power = 2916.67 J/s = 2916.67 W.

Therefore, the output power the weightlifter is 2916.67 W.

5 0

3 years ago

What is the magnitude of the velocity when the elastic potential energy is equal to the kinetic energy? (Assume that U=0 at equi

zhannawk [14.2K]

Answer:

Explanation:

General Equation of SHM is given by

$x=A\cos \omega t$

$v=-A\omega \sin \omega t$

where x=position of particle

A=maximum Amplitude

$\omega =$ angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. $\frac{1}{2}kA^2$

where k=spring constant

Potential Energy is given by $U=\frac{1}{2}kx^2$

also it is given that Potential Energy(U) is equal to Kinetic Energy(K)

Total Energy $=K+U$

Total $=2U=2\times \frac{1}{2}kx^2$

$\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2$

$x=\pm \frac{A}{\sqrt{2}}$

at $x=\frac{A}{\sqrt{2}}$

velocity is $v=\frac{A\omega}{\sqrt{2}}$

6 0

3 years ago

Which one of the following is not equivalent to 2.50 miles?

lisabon 2012 [21]

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)$

$x = 13200\,ft$

$x = 1.320\times 10^{4}\,ft$ (Choice A)

mi - in

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)$

$x = 158400\,in$

$x = 1.584\times 10^{5}$ (Choice B)

mi - km

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)$

$x = 4.025\,km$ (Different from Choice C)

mi - yd

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft} \right)$

$x = 4400\,yd$

$x = 4.40\times 10^{3}\,yd$ (Choice D)

mi - cm

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)$

$x = 402500\,cm$

$x = 4.025\times 10^{5}\,cm$ (Choice E)

Choice C is not equivalent to 2.50 miles.

6 0

4 years ago

An object that looks white when exposed to sunlight reflects all colors of light. What

Ierofanga [76]

It looks blue as it is only reflecting blue light

6 0

3 years ago