Question 15

A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical heig

kolezko [41]

Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h conservation of energy

I = 2/5 m R^2 inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

8 0

2 years ago

An analogy makes a comparison between objects based on their similar qualities. Cassidy wanted to create an analogy for the moti

user100 [1]

The three phases of matter differ in properties just because of the proximity of their molecules. The solid phase is the most organized of all. Its atoms are compactly arranged together and has the strongest intermolecular forces to keep them together. This is why they have a definite shape and volume. The liquid phase have molecules that are far away from each other, but not as far as that of the gas phase. The liquid and gas phases can be lumped into one group called fluids because they have the same property - they take the shape and volume of their container.

To make an analogy, see the attached picture for your reference.

8 0

3 years ago

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Which of the following changes will increase the period of an oscillating spring mass system?

Gnoma [55]

Answer:

a. an increase in the mass on the spring.

Explanation:

T = 2π/ω = 2π/√(k/m) = 2π√(m/k)

2π is a constant

as m is in the numerator, increasing mass will increase the period.

6 0

2 years ago

Differences between relative density and density

Reika [66]

Answer:

Density is the ratio between the mass and the volume of a body. Relative density, on the other hand, is the ratio between the density of an object (substance) and the density of some other reference object (substance) at some given temperature.

Explanation:

7 0

2 years ago

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Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

$M = \frac{gr^2}{G}$

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

$g = 9.8m/s^2\\G = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m$

Replacing at this equation we have that

$M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg$

The Volume of a Sphere is equal to

$V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3$

Therefore using the relation between mass, volume and density we have that

$\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3$

6 0

3 years ago