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3 years ago

How is volume and temperature related?

1 answer:

5 0

Charles' Law: The Temperature-Volume Law. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. As the volume goes up, the temperature also goes up, and vice-versa.

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A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

A 750-kg automobile is moving at 16.8 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car

anygoal [31]

Answer:h=19.4 m

Explanation:

Given

mass of automobile $m=750\ kg$

Initial height of automobile $h_o=5\ m$

Velocity at this instant $v=16.8\ m/s$

If the car stops somewhere at a height $h$

Thus conserving total energy we get

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2+mgh_o=\frac{1}{2}m(0)^2+mgh$

$\frac{v^2}{2g}+h_o=h$

$h=5+\frac{16.8^}{2\times 9.8}$

$h=5+14.4$

$h=19.4\ m$

6 0

3 years ago

What is the great red spot? what is the great red spot? a long-lived, high-pressure storm on jupiter a place where reddish parti

maxonik [38]

A long-lived, high-pressure storm on jupiter a place where reddish particles from io impact jupiter's surface

5 0

3 years ago

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago

Can you please answers these for me please today is the last day to turn in work and I need this to pass please I’m begging than

Ymorist [56]

Answer:

1. F = ma F = 0.2kg * 20m/s² = 4Kg * m/s² = 4N

2. F = ma F - 18Kg * 3m/s² = 54Kg * m/s² = 54N

3. F = ma F = 0.025Kg * 5m/s² = 0.125N

4. F = ma F = 50Kg * 4m/s² = 200N

5. F = ma F = 70Kg * 4m/s² = 280N

6. F = ma F = 9Kg * 9.8m/s² = 88.2N

Explanation:

Hope this helps ! ^^

8 0

2 years ago