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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:

The value is  v = 1.309*10^{5}\ m/s

Explanation:

The radius is r = 7.40 *10^{20} \  m

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Generally the speed of the star is mathematically represented as

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Here G is the gravitational constant with a value

        G = 6.67384 * 10^{-11}

So

      v = \sqrt{\frac{6.67384 * 10^{-11} * 1.90 *10^{41}}{7.40 *10^{20}} }

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3 years ago
What is the frequency of a microwave with a wavelength of 3.52 mm?
AlekseyPX

As per the question the wavelength of the microwave is given as 3.52 mm.

we are asked to calculate the frequency of the wave.

we know that microwave is a electromagnetic wave.

As per Clark Maxwell's electromagnetic theory ,every electromagnetic wave moves with a velocity equal to the velocity of light in vacuum and that is equal to 3×10^8 m/s.

From the equation of the wave,we know that velocity of wave is the product of frequency and wavelength.

Mathematically   wave velocity v=f*\lambda   where f is the frequency of the wave and \lambda is the wavelength.

As per the question \lambda=3.52 mm

                                               = 3.52*10^{-3} m

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Hence frequency of the wave f=\frac{v}{\lambda}

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3 years ago
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Answer:

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Explanation:

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Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

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Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

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(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

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