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3 years ago

If you were to measure of liters of gas trapped in a bottle, which of the following would you be describing?

Physics

2 answers:

lina2011 [118]3 years ago

8 0

Your answer is C. Volume

Ratling [72]3 years ago

4 0

Answer: volume

Explanation:

Horsepower is defined as the unit of measurement of the rate at which work is done.

Mass is defined as the amount of matter contained in the body. It is measured in units like gram(g), kilogram (kg) and milligram (mg).

Volume of the gas is defined as the space occupied by a substance. It is expressed in units like $cm^3$ , $m^3$ , Liters (L) and milliliters (ml)

Pressure of the gas is defined as the force exerted by the particles on the walls of the container. It is expressed in various terms like 'mmHg', 'atm', 'kiloPascals' etc.

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A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme

mart [117]

Answer:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

Explanation:

let $m$ be the mass attached, let $k$ be the spring constant and let $\beta$ be the positive damping constant.

-By Newton's second law:

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}$

where $x(t)$ is the displacement from equilibrium position. The equation can be transformed into:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$ shich is the equation of motion.

7 0

3 years ago

HELP!!!!

kykrilka [37]

Answer:

Clouds form when below the dew point

4 0

3 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

Horizontally:

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

Vertically:

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :