Ask question

Ask question

All categories

3 years ago

14

If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le

t's analyze the restoring force driving the floating object back to equilibrium. Suppose the cube is distance y below its equilibrium position. Find an expression for the y-component of the net force exerted on it. Simplify this expression as much as possible. Does it look like a linear restoring force?

1 answer:

GarryVolchara [31]3 years ago

8 0

Answer:

$F_{y}$ = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

B₀ - W = 0

B₀ = W

ρ_fluid g V_fluid = W

the volume of the fluid is the area of the cube times the height it is submerged

V_fluid = A y

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

B - W = m a

ρ_fluid g A (y₀ + y) - W = m a

ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

ρ_fluid g A y = m a

this equation the net force is

$F_{y}$ = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

You might be interested in

A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val

Lostsunrise [7]

Answer:

$1.25\ \mu\text{g/m}^3$

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

$\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}$

$\dot{m}$ = Mass flow rate of pollutant = 25 g/s = $25\times 10^6\ \mu\text{g/s}$

Concentration is given by

$C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3$

The steady state concentration of pollutants in the valley, is $1.25\ \mu\text{g/m}^3$ .

6 0

3 years ago

Ejercicio 1: Un cuerpo gira en un círculo de 8cm de diámetro con una rapidez constante

Lera25 [3.4K]

Answer:

Exercise 1;

The centripetal acceleration is approximately 94.52 m/s²

Explanation:

1) The given parameters are;

The diameter of the circle = 8 cm = 0.08 m

The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m

The speed of motion = 7 km/h = 1.944444 m/s

The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²

The centripetal acceleration ≈ 94.52 m/s²

8 0

3 years ago

g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positi

Arisa [49]

Answer:

$F_x = -\frac{6 C_6}{2^7}$

Attractive

Explanation:

Data provided in the question

The potential energy of a pair of hydrogen atoms given by $\frac{C_6}{X_6}$

Based on the given information, the force that one atom exerts on the other is

Potential energy μ = $\frac{C_6}{X_6}$

Force exerted by one atom upon another

$F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})$

or

$F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})$

or

$F_x = -\frac{6 C_6}{2^7}$

As we can see that the $C_6$ comes in positive and constant which represents that the force is negative that means the force is attractive in nature

5 0

3 years ago

An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move

lions [1.4K]

This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3

8 0

3 years ago

Read 2 more answers

two children sit on a merry go round. on sits 0.5 m from the center and the other sits 3 m from the center. Which child has grea

GuDViN [60]

i would say that the child with more linear speed is the cild that is 3 meters away from the center of the merry go round. because the child that is 0.5 meters from the center of the merry go round is less linear because the steering of the merry go round is started from the outer part of the merry go round so it would make more sense that the child that is 3 meters from the center of the merry go round would be more linear in speed.

hope this helps!

8 0

4 years ago