Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
The answer would most likely be A since obviously gravity weighs things down which helps the every other masses stay settled in place
<span>Answer:
So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z.
So we have x2 + 1002 = z2. Now take its derivative in terms of time to get
2x(dx/dt) = 2z(dz/dt)
So at your specific moment z = 200, x = 100âš3 and dx/dt = +8
substituting, that makes dz/dt = 800âš3 / 200 or 4âš3.
Part 2
sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get
cos a (da./dt) = -100/ z2 (dz/dt)
So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or π/6 radians.
Substitute to get
cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3)
âš3 / 2 (da/dt) = -âš3 / 100
da/dt = -1/50 radians</span>
Answer:
You wouldnt fall you would be sucked and you would lose all air supply and you lungs would pop
Explanation:
