Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Answer:
b
Explanation:
i took the quiz i think its right
Force = -kx
80N=0.15m * -k
K=-80/0.15=533.333. Spring constant
Energy=1/2kx^2
1/2*(-80/0.15)*80^2=Energy
