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Vikki [24]
3 years ago
7

Which nucleus completes the following equation?

Physics
1 answer:
ICE Princess25 [194]3 years ago
7 0

Answer:

Option D. ²³⁹₉₃Np

Explanation:

From the question given above, the following data were:

²³⁹₉₂U —> ⁰₋₁e + __

Let ⁿₘX represent the unknown. Thus, the equation above becomes

²³⁹₉₂U —> ⁰₋₁e + ⁿₘX

Next, we shall determine n, m and X. This can be obtained as follow:

239 = 0 + n

239 = n

n = 239

92 = –1 + m

Collect like terms

92 + 1 = m

93 = m

m = 93

ⁿₘX => ²³⁹₉₃X => ²³⁹₉₃Np

Thus, the balanced equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ⁿₘX

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

Option D gives the correct answer to the question.

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How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers
astraxan [27]

Answer:

18min

Explanation:

v=d/t

t=d/v= 27/90 =0.3hrs =18min

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3 years ago
What does digital media allow you to do?
kifflom [539]
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4 0
3 years ago
A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The
VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

\dfrac{1}{2}mu^2 = m g d sin \theta

u^2 = 2 g d sin30^0

u= \sqrt{2\times 9.8 \times 10 sin30^0}

u = 9.89 m/s

Using second law of motion  

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,  

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0
3 years ago
What could be used as another word for electrical potential
tekilochka [14]

Answer:

hey mate

answer is probably voltage as per me

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Explanation:

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5 0
2 years ago
Read 2 more answers
A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

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

3 0
3 years ago
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