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2 years ago

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A 45-mH inductor is connected to an ac source of emf with a frequency of 250 Hz and a maximum emf of 20 V. If the voltage across

the inductor is +20 V at time t = 0, what is the voltage at time t = 2.0 ms? Assume the phase constant is zero.

1 answer:

ruslelena [56]2 years ago

5 0

Answer:

-20v

Explanation:

Data provided in the question as per the given situation

L = 45 mH

f = 250 Hz

V_o = 20V

Based on the above information, let us assume voltage across inductor be V at t

So,

$V = 20 sin (2 \pi \times 250\times t + \frac{\pi}{2})$

t = 2 ms

Now

The voltage at time t = 2.0 ms is

$V_1 = 20 sin (500\times 2\times 10^{-1} \times \pi + \frac{\pi}{2})$

$= 20 sin(\pi + \frac{\pi}{2} )$

= -20 volt

= -20v

We simply solved the above equation so that the correct voltage could come

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An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees

vitfil [10]

Answer:

0.002925 m

Explanation:

Lt = LO(1 +α Δt ) here Lt is total length Lo is original length α is coefficient of linear expansion and Δt is change in temperature

<h2>for aluminium</h2>

α=25×10^-6

Lt = 5(1+25×10^-6×(70-20))

Lt = 5 (1+25×10^-6×50)

Lt = 5 ( 1+0.00125)

Lt = 5×1.00125

Lt =5.00625 m

<h2>for nickel </h2>

α=13.3×10^-6

Lt =5(1+13.3×10^-6×50)

Lt = 5(1+0.000665)

Lt =5.003325 m

hence difference in length =5.00625-5.003325

= 0.002925 m

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2 years ago

A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage

Alex787 [66]

The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip.

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.

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2 years ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v $\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

2 years ago

The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.

Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

$W=0.24\times 1\times 10^3=240\ W/m^2$

Energy required by the United States is $1\times 10^{19}\ J/yr$ (assumed)

Power

$P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W$

Area

$A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2$

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

$1\ m^2=\frac{1}{1000\times 1000}\ km^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2$

Converting to mi²

$1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2$

Each side of the square would be 1320.33 kilometers or 509.79 miles

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2 years ago

PLEASE ANSWER ACCURATELY DO NOT GUESS PLEASE AND THANK YOU

krek1111 [17]

Hello! I can help you with this!

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2 years ago