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4 years ago

What is the charge value of 45 electrons

Physics

1 answer:

tamaranim1 [39]4 years ago

8 0

Answer:

$7.2\times 10^{-18}C$

Explanation:

-Let x be the charge of 45 electrons

-An electron has a relatively law mas.

-Using the law of proportions, the charge value of 45 electrons is calculated as:

$1e=1.6 \times 10^{-19} coulomb\\45e=x\\\\x=\frac{45\times1.6 \times 10^{-19} coulomb}{1e}\\\\x=7.2\times 10^{-18}C$

Hence, the electron charge of 45 electrons is $7.2\times 10^{-18}C$

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The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

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$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

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4 years ago

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Answer:

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Planet Orbital Eccentricity

(Point in Orbit Closest to Sun)

measured in AU's

Mercury 0.206

Venus 0.007

Earth 0.017

Mars 0.093

Jupiter 0.048

Saturn 0.056

Uranus 0.047

Neptune 0.009

Pluto 0.248

Explanation:

link to information:

https://www.enchantedlearning.com/subjects/astronomy/glossary/Eccentricity.shtml

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2 years ago

que 2. Why do we keep frequency constant instead of keeping vibrating length constam second law of vibrating string?

ella [17]

Answer:

The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant

The law can be expressed mathematically as follows;

$f = \dfrac{1}{2\cdot l} \cdot \sqrt{\dfrac{T}{m} }$

The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously

Therefore, the law is verified indirectly, by rearranging the above equation as follows;

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Explanation:

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