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meriva
3 years ago
13

A vocalist with a bass voice can sing as low as 92 Hz.

Physics
1 answer:
Inessa05 [86]3 years ago
5 0

Answer:

  • 3.26 x 10 to the power of 6

Explanation:

c = lambda × frequency

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The bottom one has the highest amplitude
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According to the equation for the Balmer line spectrum of hydrogen, a value of n = 3 gives a red spectral line at 656.3 nm, a va
nirvana33 [79]

Answer:

E3 = 3.03 10⁻¹⁶ kJ,  E4 = 4.09 10⁻¹⁶ kJ and  E5 = 4.58 10⁻¹⁶ kJ

Explanation:

They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units

     n = 3        λ = 656.3 nm = 656.3 10⁻⁹ m

     n = 4        λ = 486.1 nm = 486.1 10⁻⁹ m

     n = 5        λ=434.0 nm = 434.0 10⁻⁹ m

Let's use the Planck equation

     E = h f

The speed of light equation

   c = λ f

replace

    E = h c /λ

Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s

Let's calculate the energies

     E = 6.63 10⁻³⁴ 3 10⁸ / λ

     E = 19.89 10⁻²⁶ /λ

n = 3

    E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹

    E3 = 3.03 10⁻¹⁹ J

    1 kJ = 10³ J

    E3 = 3.03 10⁻¹⁶ kJ

n = 4

    E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹

    E4 = 4.09 10⁻¹⁹ J

    E4 = 4.09 10⁻¹⁶ kJ

n = 5

    E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹

    E5 = 4.58 10⁻¹⁹ J

    E5 = 4.58 10⁻¹⁶ kJ

7 0
3 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
s344n2d4d5 [400]

Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
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Answer:

There are three ways an object can accelerate: a change in velocity, a change in direction, or a change in both velocity and direction.

Explanation:

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