Please List 3 categories of properties used to classify solids.

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago

A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

The height of water in the column = 348.14 cm

Explanation:

Pressure:This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²

p = Dgh............... Equation 1

Where p = pressure, D = density, g = acceleration due to gravity, h = height.

From the question, the same pressure will support the column of mercury and water.

p₁ = p₂

Where p₁ = pressure of mercury, p₂ = pressure of water

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

h₂ = 348.14 cm

The height of water in the column = 348.14 cm

6 0

3 years ago

engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi

Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

where ;

$m_1=mass of object 1\\v_1i=initial velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2$