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xz_007 [3.2K]
2 years ago
13

D

Physics
1 answer:
Veronika [31]2 years ago
5 0

Answer:

english please

Explanation:

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Matter is lost when a candle is<br> burned.<br> True or False?
Advocard [28]

Answer:

false

Explanation:

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What is the layer of rocks and moon dust called
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I believe it is called the "Lunar Crust" but not 100% sure.
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Two hoops, starting from rest, roll down identical inclined planes. The work done by nonconservative forces, such as air resiste
Maurinko [17]

Answer: both hoops have the same kinetic energy at the bottom of the incline.

Explanation:

If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.

K1 + U1 = K2 + U2

If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then

K1 = 0 and U2 = 0

⇒ ΔK = ΔU = m g. h

If both inclines have the same height, and both hoops have the same mass m, the  change in kinetic energy, must be the same for both hoops.

6 0
3 years ago
A car is traveling at 60 mph. If the rate of speed increases 4 mph each hour,
Sauron [17]

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5 hours ................

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3 years ago
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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

4 0
3 years ago
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