An Object moving in a liquid experiences a linear drag force: D= (bv, direction opposite the motion), where b is constant called
the drag coefficient. For a sphere of radius R, the drag constant can be computed as b = 6πηR, where η is the viscosity of the liquid.
a. Use what you've learned in calculus to prove that
ax= vx . dvx/dx
b. Find an algebraic expression for , Vx (x) the x-component of velocity as a funtion of distance traveled, for a spherical particle of radius R and mass m that is shot horizontally with initial speed Vo through a liquid viscosity η.
c. Water at 20°C has viscosity η= 1.0 x 10^-3 Ns/m^2.Suppose a 1.0-cm-diameter, 1.0 g marble is shot horizontally into a tank of 20°C water at 10 cm/s. How far will it travel before stopping?
a. Use what you've learned in calculus to prove that
ax= vx . dvx/dx
b. Find an algebraic expression for , Vx (x) the x-component of velocity as a funtion of distance traveled, for a spherical particle of radius R and mass m that is shot horizontally with initial speed Vo through a liquid viscosity η.
c. Water at 20°C has viscosity η= 1.0 x 10^-3 Ns/m^2.Suppose a 1.0-cm-diameter, 1.0 g marble is shot horizontally into a tank of 20°C water at 10 cm/s. How far will it travel before stopping?
1 answer:
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Answer:
The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters
Explanation:
Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;
v² = u² - 2·g·h
Where;
v = The final velocity
u = The initial velocity
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height she jumps
At the maximum height, = 0.500 m, she jumps, v = 0, therefore, we have;
0² = u² - 2·g·
u² = 2 × 9.8 × 0.5 = 9.8
u = √9.8 ≈ 3.13
u = 3.13 m/s
Her initial jumping velocity ≈ 3.13 m/s
Escape velocity,
Where;
M = The mass of the asteroid
G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
r = The radius of the asteroid
The average density of the Earth = 5515 kg/m³
The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³
The escape velocity, she has, ≈ 3.13 m/s is therefore;
Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.