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2 years ago

What remains the same in a compound regardless of the quantity?

1 answer:

8 0

The ratio btw number of atoms.
example: water H2O is the compound. No matter how much water you take (a liter, 10 cubic meters... whole swimming pool), the proportion btw hydrogen atoms and oxygen would be always 2 to 1.

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DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY

marishachu [46]

When a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star.This explosion made waves in space which squeezed the cloud of gas & dust.

8 0

2 years ago

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she

tester [92]

Answer:

a) I = 1,75 10-² kg m² and b) I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

I = I core + I shell

The moment of inertia of a solid sphere is

I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

R = d / 2

R= 0.196 m / 2 = 0.098 m

I core = 2/5 1.6 0.098²

I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

R = 0.206 / 2

R = 0.103 m

I shell = 2/3 1.6 0.103²

I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

I = I core + I shell

I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

I = 17.47 10⁻³ kg m²

I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

R = 0.216 / 2

R = 0.108 m

It is a uniform sphere

I = 2/5 M R²

I = 2/5 3.2 0.108²

I = 1.49 10⁻² kg m²

7 0

2 years ago

Each of the following statements is arguably true of thermometers. Which of them is most helpful to keep in mind if you are cond

Lera25 [3.4K]

Answer:

The temperature reported by a thermometer is never precisely the same as its surroundings

Explanation:

In this experiment to determine the specific heat of a material the theory explains that when a heat interchange takes place between two bodies that were having different temperatures at the start, the quantity of heat the warmer body looses is equal to that gained by the cooler body to reach the equilibrium temperature. <u>This is true only if no heat is lost or gained from the surrounding.</u> If heat is gained or lost from the surrounding environment, the temperature readings by the thermometer will be incorrect. The experimenter should therefore keep in mind that for accurate results, the temperature recorded by the thermometer is similar to that of the surrounding at the start of the experiment and if it differs then note that there is either heat gained or lost to the environment.

3 0

3 years ago

A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface

Fiesta28 [93]

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$ : The energy when the first spring is compress

2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3. $E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so: