If a school bus moves with the speed of 72km/hr, how many meters does it cover in one hour?
2 answers:
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<span>The correct option is D) both A and B. In fact, the strength of an electric field produced by a single point charge is given by
</span>
<span>
where Q is the charge that generates the field, k is the Coulomb's constante and r is the distance from the charge source of the field.
From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>
</span>
where Q is the charge that generates the field, k is the Coulomb's constante and r is the distance from the charge source of the field.
From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
(a) 50 rad/s
The angular speed of the CD is related to the linear speed by:
where
is the angular speed
v is the linear speed
r is the distance from the centre of the CD
When scanning the innermost part of the track, we have
v = 1.25 m/s
r = 25.0 mm = 0.025 m
Therefore, the angular speed is
(b) 21.6 rad/s
As in part a, the angular speed of the CD is given by
When scanning the outermost part of the track, we have
v = 1.25 m/s
r = 58.0 mm = 0.058 m
Therefore, the angular speed is
(c) 5550 m
The maximum playing time of the CD is
And we know that the linear speed of the track is
v = 1.25 m/s
If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:
(d)
The angular acceleration of the CD is given by
where
is the final angular speed (when the CD is scanned at the outermost part)
is the initial angular speed (when the CD is scanned at the innermost part)
is the time elapsed
Substituting into the equation, we find