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
The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.
<h3>Resistance</h3>Resistance is a measure of the opposition to flow of electric current. It is measured in ohms.
It is given by the formula:
Given that R₂ = 469 ohm, hence:
But l₁ = 6l₂, r₁ = (1/5)r₂, hence:
The equivalent resistance (R) is:
The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.
Find out more on resistance at: brainly.com/question/17563681
Answer:
the work done by gravity on the boy is 604.62 J
Explanation:
Given;
distance the boy slides, d = 3 m
angle of inclination of the playground, θ = 40⁰
mass of the boy, m = 32 kg
The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.
The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.
The work done by gravity on the boy is calculated as;
W = P.E = mgh
= 32kg x 9.8m/s² x 1.928m
= 604.62 J
Therefore, the work done by gravity on the boy is 604.62 J
<u>Answer</u>:
The coefficient of static friction between the tires and the road is 1.987
<u>Explanation</u>:
<u>Given</u>:
Radius of the track, r = 516 m
Tangential Acceleration = 3.89 m/s^2
Speed,v = 32.8 m/s
<u>To Find:</u>
The coefficient of static friction between the tires and the road = ?
<u>Solution</u>:
The radial Acceleration is given by,
Now the total acceleration is
=>
=>
=>
=>
The frictional force on the car will be f = ma------------(1)
And the force due to gravity is W = mg--------------------(2)
Now the coefficient of static friction is
From (1) and (2)
Substituting the values, we get