Answer:
He is warmed up now
Explanation:
His muscles are better and stretched now
Explanation:
mass, m = 5kg
initial velocity, u = 16m/s
final velocuty, v = -22m/s
change in momentum, ∆p = ?
∆p = m (v-u)
5(-22-16)
5(38)
∆p = 190kgm/s
check the calculations!
The answer is B. The obvious questions do not always lead to interesting results. This is because if everyone did the same thing every time, we wouldn't find any new data. With variables, we have so many different options, leading to creativity and new data found!
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
Learn more about displacement here: brainly.com/question/321442
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Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest