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4 years ago

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Which law is associated with inertia? *

1 answer:

defon4 years ago

4 0

Newton first law of motion

Explanation:

This law by Newton states that an object will remain at rest, or a moving object will stay in motion in a straight line and with the same speed unless an external and unbalanced force acts on the object.

One application of this Newtonian law is in the launching of projectiles like ballistic missiles. A ballistic missile parabolic path is affected by its inertia and the unbalanced forces acting on the object such as air resistance/ drag and gravity.

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The kinetic molecular theory states that all particles of an ideal gas are

Marizza181 [45]

THE KINETIC MOLECULAR THEORY STATES THAT ALL PARTICLES OF AN IDEAL GAS ARE IN CONSTANT MOTION AND EXHIBITS PERFECT ELASTIC COLLISIONS.

Explanation:

An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.

The kinetic-molecular theory as it applies to gases has five basic assumptions.

Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size.
Gas particles are in constant rapid motion in random directions.
Collisions between gas particles and between particles and the container walls are elastic collisions.
The average kinetic energy of gas particles is dependent upon the temperature of the gas.
There are no forces of attraction or repulsion between gas particles.

5 0

3 years ago

VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu

SOVA2 [1]

Answer:

31.9 °C

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹

ΔT = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

6 0

3 years ago

Can sugar and water be separated using distillation?

Bogdan [553]

Answer:

Yes.

Explanation:

Because distillation separates substances based on their different boiling points.

8 0

3 years ago

Read 2 more answers

Which of the following is an element?

Kryger [21]

Oxygen is a chemical element with symbol O and atomic number 8. Classified as a nonmetal, Oxygen is a gas at room temperature.

8 0

3 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago