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Arturiano [62]
3 years ago
5

A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5​​​​m​​/s2 for 4s. After t

he wind stops, the sailboat is traveling to the left with a velocity of 3.0​m/s.
Assuming the acceleration from the wind is constant, what was the initial velocity of the sailboat before the gust of wind?
Answer using a coordinate system where rightward is positive.
Physics
2 answers:
Rus_ich [418]3 years ago
4 0

Explanation :

Let rightward direction is considered as positive and leftward direction is considered as negative.

Given that,

Acceleration of the boat, a=-2.5\ m/s^2 ( in left )

Time taken, t=4\ s

Final velocity of the boat, v=-3\ m/s

We have to find initial velocity (u) of the sailboat.

Using first equation of motion :

v=u+at

u=v-at

u=-3-(-2.5)\times 4

u=-13\ m/s

So, initially the boat is moving with a speed of 13 m/s in leftwards direction.

Hence, this is the required solution.

Pavlova-9 [17]3 years ago
3 0
Taking right movement to be positive means leftward movement is negative.
Hence we have a deceleration of
a = - 2.5 {ms}^{ - 2}

t = 4s
v = 3.0 {ms}^{ - 1}
Using this 'suvat' equation
v = u + at
we can determine the initial velocity

3.0= u + -2.5(4)

3.0+2.5(4)=u

13.0 = u

Hence the initial velocity is 13.0 meters per seconds
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A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
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Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

E = 10r, E = 10t

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