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Andrei [34K]
2 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]2 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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Answer:

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Explanation:

Given;

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initial temperature of the copper cube, T_c_u = 85°C

initial temperature of the aluminum chunk T_A_l = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}

where;

C_{AL} is specific heat capacity of aluminum

\delta T_{Al} is change in temperature of aluminum

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\delta T_c_u is the change in temperature of copper

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg

Therefore, the mass of the aluminum chunk is 258 g

7 0
3 years ago
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Answer:

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7 0
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