Answer:
(a) 1.21 m/s² (b) 1.75 m/s²
Explanation:
The initial speed of the car, u = 17.8 m/s
Case 1.
Final speed of the car, v = 23.5 m/s
Time, t = 4.68-s
Acceleration = rate of change of velocity
Case 2.
Final speed of the car, v = 15.3 m/s
Hence, this is the required solution.
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
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Answer: -33.3 * 10^9 C/m^2( nC/m^2)
Explanation: In order to solve this problem we have to use the gaussian law, the we have:
Eoutside =0 so teh Q inside==
the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.
Then we simplify and get
σ= -4.6/(2*π*b)= -33.3 nC/m^2
