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3 years ago

What are three physical properties of gases

1 answer:

7 0

Answer: Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form. An internal combustion engine provides a good example of the ease with which gases can be compressed.

Explanation:

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15. What is a thermograph?

Genrish500 [490]

Answer: A device that uses infrared sensors.

Explanation:

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3 years ago

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How fast would 40 Newtons of force accelerate a 2 kg object?

Digiron [165]

Answer:

$20 m/s^2$

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force on the object

m is its mass

a is its acceleration

In this problem:

F = 40 N is the force on the object

m = 2 kg is its mass

Therefore, the acceleration of the object is

$a=\frac{F}{m}=\frac{40}{2}=20 m/s^2$

8 0

3 years ago

How does kinetic energy connect or relate to Newtons second law of motion?

Orlov [11]

Newtons second law of motion: "T<span>he acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."

kinetic energy is energy that an object posses while in motion and to get that it must have potential energy.</span>

8 0

3 years ago

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A gas in a sealed container has a pressure of 50 kPa at 27°C. What will the pressure of the gas be if the temperature rises to 8

alexgriva [62]

Answer:

the final pressure of the gas is 60 kPa.

Explanation:

Given;

initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa

initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k

final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K

Let the final pressure of the gas = P₂

Apply pressure law;

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} = \frac{50,000 \times 360}{300} = 60,000 \ Pa = 60 \ kPa$

Therefore, the final pressure of the gas is 60 kPa.

4 0

3 years ago

A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

$\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s$

4 0

3 years ago