Answer:
The answer is below
Explanation:
Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.
The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph
vₐ = vₙ + vₐₙ
vₐ = vₐₙ
Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.
The direction of the wind θ is:
sin(θ / 2) = vₙ / 2vₐ
sin(θ / 2) = 70/ (2*135)
sin(θ / 2) = 0.2593
θ / 2 = sin⁻¹(0.2593) = 15
θ = 30⁰
2α = 180° - 30°
2α = 150°
α = 75°
a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.
Answer:
a)v= 675V
b) v=-675V
c)v=-450V
Explanation:
Formula: V = k*sum(qi/ri)
Given:
q1 = +6.00nC --> R1 = 2.00cm
q2 = -9.00nC --> R2 = 4.00cm
(a) r = 0. So, inside both shells
V=675V
(b) r = 4.00cm. So, This point is outside of shell 1 but inside shell 2.
V=-675V
(c) r = 6.00cm. So, This point out of both shells.
there it is fella tried on ma own consciousness
Answer:
- Newton's first law applies. An object at rest will stay that way until a force is applied.
- Any amount of effort can be applied to any amount of mass (in the ideal case). The question is not sufficiently specific.
Explanation:
A force is required to move an object because the object will stay at rest until a force is applied.
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The effort required to lift or push two masses instead of one depends on the desired effect. For the same kinetic energy, no more effort is required. For the same momentum, half the effort is required for two masses. For the same velocity, double the effort is required.