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4 years ago

The equation for gear ratio

1 answer:

4 0

In a gear train with two gears, the gear ratio is defined as follows
$R= \frac{\omega _A}{\omega _B}$

where $\omega _A$ is the angular velocity of the input gear while $\omega _B$ is the angular velocity of the output gear.

This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:
$\omega = \frac{v}{r}$
But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
$R= \frac{\omega _A}{\omega _B} = \frac{r_B}{r_A} = \frac{N_B}{N_A}$

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Answer:

150000000

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49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

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Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

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4 years ago

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Answer:

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