Just about anything you could ask about a projectile AFTER it's launched
depends on both components of the launch velocity.
Here are some that I can think of:
-- angle of launch
-- magnitude of launch velocity
-- location at any time after launch
-- magnitude of velocity at any time after launch
-- direction of velocity at any time after launch
-- distance of the landing point from the launch point
The effective height of the water for Smith's house will be 24.61m.
<h3>How to calculate the height?</h3>
Based on the information given, the volume of the water in sphere will be:
= 4/3πr³ = (5.80 × 10^5)/1000
= 4.18r³ = 580
r³ = 138.7
r = 5.18m
The effective height of the water will be:
= 18.0 + 2(5.18)
= 28.36
The gauge pressure at Faucet of Jones house will be:
= pgh
= 1000(9.8)(28.36)
= 277.9kPa
The effective height of the water for Smith's house will be:
= 18.0 + 2(5.18) - 3.75
= 24.61m
The gauge pressure at Faucet of Jones house will be:
= 1000 × 9.8 × 24.61
= 241.2kPa
Learn more about height on:
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Bernoulli's principle of laminar/lamellar air flow, I think. High flow speed = low pressure, low flow speed = high pressure I think. So, the wings/aerofoils are designed to induce a low pressure on the top side of the wing and a high pressure on the underside of the wing, thus producing an "aerodynamic upthrust" (a static upthrust comes from an object in water via Archimedes) and LIFT.
Two "particles" of air one going topside and the other underside meet again at the end of their motion across the wing. So, top side has to travel faster than bottom side. So top side has a lower "dynamic pressure" than underside.
And all that for 5 points ????????? (If I'm right, of course ... )
Answer:
All kinds of waves have the same fundamental properties of reflection, refraction, diffraction, and interference, and all waves have a wavelength, frequency, speed, and amplitude.
Explanation:
Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.
