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Leya [2.2K]
3 years ago
8

1) Why does an electron have no quarks?

Physics
1 answer:
Nataliya [291]3 years ago
5 0
Electrons have electrical magnetic fields that require them to have energy that is too intense for quarks
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The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency diffe
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Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?

B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?

Answer:

The frequency change percentage is 9.94%

Explanation:

The frequency shift can be calculated as follows.

F= \frac{V+V_{0}}{V+V_{s}}

   = 9.05GHz\times\frac{343m/s+0}{343m/s+(-31.3m/s)}

   =9.95 GHz

So the frequency change seen by the detector is 9.95 - 9.05

% difference= \frac{0.9}{9.05} \times100

                    = 9.94%

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Which of the following is group 1? a. alkali metals b. alkaline earth metals c. halogens d. transition metals
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Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

Explanation:

(3) The period of a satellite is given as;

T = 2\pi \sqrt{\frac{r^3}{GM} }

where;

T is the period of the satellite

M is mass of Earth

r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

 

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N

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