Physics, 03.04.2020 22:37 yollo74

The mass of a string is 5.40 10-3 kg, and it is stretched so the tension in it is 215 N. A transverse wave traveling on this string has a frequency of 260 Hz and a wavelength of 0.60 m. What is the length of the string

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Answer from: giuliabril

= 7.56 * 10^-9 W

Explanation:

1) v = √(T/m)

v is the velocity of a wave on the string

T is the Tension

m is the mass of a unit length ( mass of a 1 m length of the string)

v = f.λ

√(T/m) = fλ

√(195/m) = 260 * 0.6

we don't know m because we don't know the length

√(195 / m) = 156

195/m = 156²

195 / m = 24,336

m = 195 / 24,336

m = 0.008 kg/m

now we know the mass of a unit length

m = M / l

M is the mass of the string

l is the length of the string

0.008 = 2.8 * 10^-3 / l

l = 2.8 * 10^-3 / 8 * 10-3

l = 2.8 / 8

l = 0.35 m

OK that's the answer, hope you understood.

2) Power on a unit area is 3.6 * 10^-6 W/m²

so to find the power on a specific area, just multiply the area by the sound intensity

assuming the ear's flat area is the specified value

power intercepted by ear = (3.6 * 10^-6 W/m²) *( 2.1 * 10^-3 m²)

= 7.56 * 10^-9 W

Answer from: Quest

the aurora is formed when charged particles emitted from the sun during a solar flare penetrate the earth's magnetic shield and collide with atoms and molecules in our atmosphere. these collisions result in countless little bursts of light, called photons, which make up the aurora.

auroras are generated by charged solar particles hitting molecules in earth's atmosphere. this usually happens in high latitudes, because our planet’s magnetic field tends to funnel these particles toward the poles. that’s why it can be experienced in northern ohio.

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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