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siniylev [52]
3 years ago
10

A boat moves ar 10.0 m/s relative to

Physics
1 answer:
Charra [1.4K]3 years ago
5 0

Answer: 208.3 s

Explanation:

Hi!

You need to calculate the velocity of the boat relative to the shore, in each trip.

Relative velocities are transformed according to:

V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\

Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s

In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s

In upstream trip we have:

V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms

In dowstream we have:

V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms

In both cases the distance travelled is 1000m. Then the time it takes the round trip is:

T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s

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The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

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I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

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What is the threshold frequency ν0 of cesium? note that 1 ev (electron volt)=1.60×10−19 j. express your answer numerically in he
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The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

\large {\boxed {E = h \times f}}

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem!

<u>Given:</u>

Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule

<u>Unknown:</u>

fo = ?

<u>Solution:</u>

\Phi = h \times fo

3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo

fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )

\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544
  • Photoelectric Threshold Wavelength : brainly.com/question/10015690

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

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