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kap26 [50]
2 years ago
5

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,

Physics
1 answer:
posledela2 years ago
8 0

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

<h3>How can the factors of the wave be found?</h3>

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = <u>4</u>

c. The signal is <u>even</u>, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

\frac{1}{2}  \cdot  \mu^{2} \cdot \omega ^{2}  \cdot \:  {a}^{2}  \times  \lambda

Which gives;

E = 0.5 × 1.571² × 1² × 4 = <u>4.935 J/kg</u>

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = <u>1.234 W/</u><u>kg</u>

Learn more about waves here:

brainly.com/question/14015797

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ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

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ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

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The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

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The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
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Answer:

Time period, T=3.05\times 10^{-5}\ s

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

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Time period is given by :

T=\dfrac{1}{32768}

T=3.05\times 10^{-5}\ s

So, the time period of the crystal's motion is 3.05\times 10^{-5}\ s. Hence, this is the required solution.

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