The displacement of Edward in the westerly direction is determined as 338.32 km.
<h3>What is displacement of Edward?</h3>
The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.
The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰
The displacement is the side of the triangle facing 150⁰ = R
R² = a² + b² - 2abcosR
R² = 150² + 200² - (2x 150 x 200)xcos(150)
R² = 62,500 - (-51,961.52)
R² = 114,461.52
R = 338.32 km
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Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
Answer:
585×10⁸ m
Explanation:
Distance = rate × time
d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)
d = 585×10⁸ m
Answer:
The speed of the block is 4.96 m/s.
Explanation:
Given that.
Mass of block = 1.00 kg
Spring constant = 500 N/m
Position 
Coefficient of friction = 0.350
(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less
Using formula of kinetic energy and potential energy
Put the value into the formula
Hence, The speed of the block is 4.96 m/s.
