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ra1l [238]
3 years ago
12

"what is the horse's average velocity during the first ten seconds of its gallop

Physics
1 answer:
vladimir1956 [14]3 years ago
6 0

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, from the completed question, determine the distance (in meters) the horse covered in the first ten seconds of it's gallop and apply the formula below.

Average velocity = distance (in meters) ÷ time (in seconds as provided in this question)

The unit for velocity (to be used here) is m/s or ms⁻¹

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By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what
Vinil7 [7]

Explanation:

1. Mass of the proton, m_p=1.67\times 10^{-27}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}

V=\dfrac{h^2}{2q_pm_p\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}

V = 45.83 volts

2. Mass of the electron, m_p=9.1\times 10^{-31}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}

V=\dfrac{h^2}{2q_em_e\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}

V=6.92\times 10^{34}\ V

V = 84109.27 volt

Hence, this is the required solution.

7 0
4 years ago
What did Thomson contribute to the model of atom
Ganezh [65]
<h3><u>Answer;</u></h3>

<em>Electrons </em>

<h3><u>Explanation;</u></h3>
  • <em><u>Thomson contributed to the model of an atom by discovery of </u></em><em><u>electrons </u></em><em><u>and thus proving the existence of sub-atomic particles in an atom. </u></em>
  • <u><em>Thomson used cathode ray tube, and demonstrated that cathode rays were negatively charged.</em></u> According to his model normally known as the plum pudding in which he stated that an atom is composed of electrons as subatomic particles that are surrounded by positive charges to balance the electrons.
5 0
4 years ago
Read 2 more answers
A high power line carries a current of 1.0 kA. What is the strength of the magnetic field this line produces at the ground, 10 m
solmaris [256]

Answer:

The strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

Explanation:

From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:

           

B = \frac{\mu_{0}I}{2\pi r} (1)      

                                     

Where \mu_{0} is the permiability constant, I is the current and r is the distance from the wire.    

             

Notice that it is necessary to express the current, I, from kiloampere to ampere.

I = 1.0kA \cdot \frac{1000A}{1kA} ⇒ 1000A

Finally, equation 1 can be used:

B = \frac{(4\pi x10^{-7}T.m/A)(1000A)}{2\pi (10m)}    

           

B = 2x10^{-5}T    

Hence, the strength of the magnetic field that the line produces is 2x10^{-5} Tesla.

         

8 0
3 years ago
If the dartboard below is used to model an atom, which subatomic particles would be located at Z? *
Anna11 [10]

I know that protons and neutrons are located at the center of an atom, so the correct answer is D

7 0
4 years ago
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J
Lyrx [107]

Explanation:

Given that,

Wavelength of the light, \lambda=4170\ A=4170\times 10^{-10}\ m

Work function of sodium, W_o=4.41\times 10^{-19}\ J

The kinetic energy of the ejected electron in terms of work function is given by :

KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J

The formula of kinetic energy is given by :

KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s

Hence, this is the required solution.

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