15 PTS! HELP! BRAINLIEST!
1 answer:
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Answer:
Explanation:
From the given information:
The coordinate axis is situated in the east and north direction.
So, the north will be the y-axis and the east will be the x-axis
Similarly, the velocity of the plane in regard to the air in the coordinate system will be
where:
= velocity of the plane in regard to the air
v = velocity
θ = angle of inclination of the plane with respect to the horizontal
replacing v = 180 km/ and θ = 20° in above equation, then:
The velocity of the airplane in the coordinate system as:
where;
= velocity of the airplane
= velocity
∅ = angle of inclination with regard to the base axis;
Then; replacing = 150 km/h and ∅ = 30°
Therefore, the velocity of the plane in the system is :
--- (1)
( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j
(-39.24 km/h)i + (13.44 km/h) j
The magnitude is:
= 41.48 km/h
The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.
The angle of motion is:
tan θ = 39.24/13.44
tan θ = 2.9
θ =
θ = 70.97°
The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.
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