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2 years ago

Define and describe the following terms: amplitude, wavelength, frequency, period, tension.

Physics

1 answer:

maw [93]2 years ago

7 0

Answer:

Amplitude—distance between the resting position and the maximum displacement of the wave

Frequency—number of waves passing by a specific point per second

Period—time it takes for one wave cycle to complete

wavelength λ - the distance between adjacent identical parts of a wave, parallel to the direction of propagation.

Tension - described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar one-dimensional continuous object, or by each end of a rod, truss member, or similar three-dimensional object

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To start a great night of doing physics homework, you sit down to pour yourself a good cup of coffee. Your coffee mug has a mass

astra-53 [7]

Answer:

1) The temperature will be closer to water

2) T = 68.239°C

3) T = 72.142°C

4) T = 69.266 °C

Explanation:

The temperature will be closer to water because the heat capacity of water > heat capacity of coffee.

137(1.089)(T - 23.8) = 137(4.186)(79.8 - T)

⇒(1.089)(T - 23.8) = (4.186)(79.8 - T)

⇒1.089 T - 25.9182 = 334.0428 - 4.186 T

⇒1.089 T + 4.186 T = 334.0428 + 25.9182

⇒5.275 T = 359.961

⇒ T = 68.239°C

137(1.089)(T - 23.8) = 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934 = 75159.63 - 941.85 T

⇒149.193 T + 941.85 T = 75159.63 + 3550.7934

⇒1091.043 T = 78710.4234

⇒ T = 72.142°C

137(1.089)(T - 23.8) + 11.7(4.186)(T - 5.2)= 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) + 48.9762(T - 5.2) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934 + 48.9762 T - 254.67624= 75159.63 - 941.85 T

⇒149.193 T + 941.85 T + 48.9762 T = 75159.63 + 3550.7934 + 254.67624

⇒1091.043 T + 48.9762 T = 78710.4234 + 254.67624

⇒1140.0192 T = 78965.09964

⇒ T = 69.266 °C

6 0

3 years ago

Which of the following choices is an example of an allele?

VikaD [51]

Answer:

Blue eye-color gene

7 0

2 years ago

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the di

vekshin1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi (\frac{D}{2})^2$