The pilot of an airplane traveling 45 m s wants to drop supplies to flood victims isolated on a patch of land 160 m below. The s
upplies should be dropped when the plane is how far from the island?
1 answer:
Answer:
257.13 m from the island
Explanation:
Using the equations of motion
uᵧ = vertical component of the initial velocity = 0 m/s
g = 9.8 m/s²
y = 160 m
Time of flight = ?
y = uᵧt + gt²/2
160 = 0 + 4.9t²
t² = 32.65
t = 5.714 s
Note uₓ = horizontal component of the initial velocity = 45 m/s
Range of/horizontal distance covered by the dropped package = uₓ t = 45 × 5.714 = 257.13 m
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Answer:
A) x and y components of the electric field (Ep) at the origin.
Epx = -1620.5 N/C
Epy = -1620.5 N/C
Ep = (1620.5(-i)+1620.5(-j)) N/C
B) Magnitude of the electric field (Ep) at the origin.
Ep= 2291.7 N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
Data
K= 9x10⁹N*m²/C²
q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C
d₁=d₂=0.190m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁ : Electric Field at point P (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.
Ep₂: Electric Field at point P (x=0, y=0) due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
Ep: Total field at point P due to charges q₁ and q₂.
Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂
Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C
Look at the attached graphic :
Epx = Ep₁= -1620.5 N/C
Epy = Ep₂= -1620.5 N/C
A) x and y components of the electric field (Ep) at the origin.
Ep = (1620.5(-i)+1620.5(-j)) N/C
B) Magnitude of the electric field (Ep) at the origin.
