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ivann1987 [24]
3 years ago
15

If you slosh the water back and forth in a bathtub at the correct frequency, the water rises first at one end and then at the ot

her. Suppose you can make a standing wave in a 157 cm long tub with a frequency of 0.3165 Hz. What is the velocity of the water wave? Answer in units of m/s.
Physics
1 answer:
notka56 [123]3 years ago
7 0

Answer:

Velocity(v) = frequency(f) × wavelength

f = 0.3165

Wavelength = 2×length(L)

L = 157cm

Convert the length in centimetres to metre = 1.57m

v = 2×1.57 × 0.3165

v = 0.99m/s

Approx. 1m/s

Explanation:

The velocity of a wave is the product of its frequency and it's wavelength. The frequency is already known. The wavelength is the distance between two successive wave crests which is formed by sloshing water back and forth in the bath tub. Sloshing water to one end of the tub will produce a wave crest first at that end then the other completing a cycle. The wavelength will be twice the length of the bath tub as it is the distance that both crests are formed.

Wave crest is the highest point of a wave, and in this case is where the water rises to a high point in the bath tub

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Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

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6 0
2 years ago
An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports
finlep [7]

Option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

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Option D satisfies the fill-in blanks option.

Hence option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>

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6 0
2 years ago
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
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Answer:

6787.5 V

Explanation:

From the question,

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make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

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zhuklara [117]

Answer:

metal> metalloids >nonmetals    (Electrical conductivity)

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Electrical conductivity of objects can be compared by the bonding energy of electrons in them.

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Nonmetals have a very high bonding energy of electrons, so  at room temperature negligible number of free electrons are present so electrical conductivity is very low.

Metalloids have both metallic and non metallic features. The electron bonding energy falls in between that of metals and nonmetals. So electrical conductivity also lies in between metals and nonmetals.

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