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professor190 [17]

2 years ago

10

What is the velocity of a wave with a wavelength of 9 meters and a period of 0.006

1 answer:

Ipatiy [6.2K]2 years ago

8 0

Answer: 1,500m/s

Explanation:

Relationship existing between velocity of a wave (v), wavelength(¶) and frequency(f) is

v = f¶... (1)

Since Frequency (f) is the reciprocal of the period (T);

Frequency = 1/Period i.e F = 1/T... (2)

Substituting equation 2 into 1 we have;

v = 1/T × ¶

v = ¶/T

Given wavelength ¶ = 9m

Period T = 0.006s

v = 9/0.006

v = 1,500m/s

The velocity of the wave will be 1,500m/s

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The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid

faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

$w=2250*2\pi *\frac{1}{60}\\ w=235.61$

$I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}$

Now using Inertia an w

$E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J$

average power= $\frac{11.4J}{0.230s}=50.2 W$

b).

power=t*w

P= $11.5465*0.25*235.61$

P=2.957 kW

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3 years ago

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Answer:c

Explanation:

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3 years ago

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

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In order to find the electric flux we first have to find out the area of triangle.

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Electric Flux = E. A

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= 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

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3 years ago

This company is run by kids right

GREYUIT [131]

Answer:

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An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta

icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

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$v=\sqrt{\frac{T}{\mu} }$

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$v=\sqrt{\frac{200}{1\times 10^{-3} }$

$v=447.2 m/s$

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3 years ago