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2 years ago

Which of these experiments would make use of qualitative data?

Physics

1 answer:

pychu [463]2 years ago

7 0

"A study of the different amounts of time it takes water to evaporate completely"

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A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e

jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?$

Putting values in above equation, we get:

$\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K$

Hence, the final temperature is 350 K

8 0

3 years ago

You Discover A Beautiful Island Upon Which You May Build Your Own Society. You Make The Rules. What Is The First Rule You Put In

alekssr [168]

Answer: no plastic

Explanation:

4 0

2 years ago

Read 2 more answers

At standard temperature and pressure, carbon dioxide has a density of 1.98 kg/m3. What volume does 1.70 kg of carbon dioxide occ

nikitadnepr [17]

Answer:

<h2>volume= 0.85m^3</h2>

Explanation:

<em>The density of a substance is defined as the mass per unit volume of the substance, the unit is in kg/m^3 and it is represented by the greek letter rho</em>

Step one:

given data

we are told that the density of Co2= 1.98 kg/m3

and the mass of Co2 is= 1.70 kg

we know the relation between mass, volume and density is

$density=mass/volume$

make volume subject of formula we have

$volume=mass/density$

substitute we have

$volume=1.7/1.98\\\\volume= 0.85m^3$

8 0

3 years ago

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

3 0

3 years ago

Which will increase the kinetic energy of an object more, increasing its mass or increasing its speed? Why?

agasfer [191]

Increasing its velocity will add to the kinetic energy more as the formula for kinetic energy is 0.5*m*v^2. (The speed will be squared making it greater)

4 0

3 years ago