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Andrei [34K]
3 years ago
5

A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p

eriod in milliseconds, (c) amplitude in volts, and (d) phase angle in degrees
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) f=1617.9\ Hz

(b) T=0.618\ ms

(c) A=20 \ Volts

(d) \varphi=60^o

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

V(t)&=A\cdot \sin(2\pi ft+\varphi )

Where

A=Amplitude

f=frequency

\varphi=Phase\  angle

The voltage of the question is the sinusoid expression  

V(t)=20cos(5\pi\times 103t+60^o)

(a) By comparing with the general formula we have

f=5\pi\times 103=1617.9\ Hz

\boxed{f=1617.9\ Hz}

(b) The period is the reciprocal of the frequency:

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

\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec

Converting to milliseconds

\boxed{T=0.618\ ms}

(c) The amplitude is

\boxed{A=20 \ Volts}

(d) Phase angle:

\boxed{\varphi=60^o}

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A piece of glass has a temperature of 72.0 degrees Celsius. The specific heat capacity of the glass is 840 J/kg/deg C. A liquid
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Answer:

741 J/kg°C

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Final temperature, T(2) = 57° C

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Using the relation

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Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
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For maximum range we can write

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2 D =u₃² sin2θ₃ / g

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1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

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8 0
3 years ago
Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

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Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

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Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

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Substitute the values in above equation

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Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

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