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3 years ago

I’m not sure how to solve this

Physics

1 answer:

spayn [35]3 years ago

6 0

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

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3 years ago

Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave det

alekssr [168]

Answer:

1.5106 cm

Explanation:

The beat frequency is equal to the absolute value of the difference between the frequencies of the two signals:

$f_B = |f_1 - f_2|$

using the wave equation, we can re-write each frequency as

$f=\frac{c}{\lambda}$

where c is the speed of light and $\lambda$ is the wavelength. Therefore,

$f_B = |\frac{c}{\lambda_1}-\frac{c}{\lambda_2}|$

where:

$f_B = 140 MHz = 140\cdot 10^6 Hz$ is the beat frequency

$\lambda_1 = 1.50 cm = 0.015 m$ is the wavelength of the first generator

$\lambda_2$ is the wavelength of the second generator

We also know that the second generator emits the longer wavelength, so we already know that the term inside the module is positive. Therefore, we can now solve for $\lambda_2$ :

$f_B = c(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})\\\lambda_2=(\frac{1}{\lambda_1}-\frac{f_B}{c})^{-1}=(\frac{1}{0.015}-\frac{140\cdot 10^6}{3\cdot 10^8})^{-1}=0.015106 m = 1.5106 cm$