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2 years ago

12

Explain what happens to the gas pressure in a balloon as the temperature decreases

1 answer:

Julli [10]2 years ago

5 0

Answer: The pressure decreases as the temperature decreases.

Explanation: As the temperature decreases, the particles of gas start moving more slowly. The particles' velocity decreases and as it collides with the wall of the container, it exerts a less force on the walls of the container. This follows on from the equation Ft = mv - mu. Less force therefore means less pressure as force and pressure are proportional according to equation Pressure = Force / Area.

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2. An object with moment of inertia ????1 = 9.7 x 10−4 kg ∙ m2 rotates at a speed of 3.0 rev/s. A 20 g mass with moment of inert

svetoff [14.1K]

Answer:

2.85 rad/s

Explanation:

5 cm = 0.05 m

20 g = 0.02 kg

When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:

$I_2 = 1.32\times10^{−6} + 0.02 * 0.05^2 = 0.513\times10^{-4} kgm^2$

So the total moment of inertia of the system of 2 objects after the drop is:

$I = I_1 + I_2 = 9.7\times10^{-4} + 0.513\times10^{-4} = 0.0010213 kgm^2$

From here we can apply the law of angular momentum conservation to calculate the post angular speed

$\omega_1 I_1 = \omega_2 I$

$\omega_2 = \omega_1 \frac{I_1}{I} = 3 \frac{9.7\times10^{-4}}{0.0010213} = 2.85 rad/s$

6 0

3 years ago

What is a prediction

faust18 [17]

An educated guess about something. (What might happen in the future)

7 0

3 years ago

Read 2 more answers

Thorium-232 decays by alpha decay. Part of the nuclear equation is shown below. Fill in the blank with a number. 232/90

UkoKoshka [18]

its 228 on my apex......

8 0

3 years ago

If a wave has a frequency of 74 Hz and a speed of 444 m/s, what is the wave's wavelength?

DaniilM [7]

Answer:

6 m

Explanation:

velocity = wavelength x frequency

444 m/s = wavelength x 74 Hz

444 m/s / 74 Hz = wavelength

wavelength = 6m

4 0

2 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago