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Verizon [17]
3 years ago
5

We had an experiment that is entitled "balloon rocket" it is an experiment using the force of balloon to make the straw move. it

is about law of interaction.
Did the escaping air have to hit anything to make the balloon move??

my answer here is no, because even in space, space rockets can travel even in a vacuum.

is it correct??
Physics
1 answer:
andreev551 [17]3 years ago
3 0
Air from a balloon does not need to touch anything in order for the balloon to move. The outside forces around the balloon push down it so it contracts and air is expelled. This propulsion is what moves the balloon.
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A rigid container holds 0.30g of hydrogen gas.
Strike441 [17]

Answer:

Part A:    \mathbf{Q =94 \ J} to two significant figures

Part B:    \mathbf{Q =160  \ J} to two significant figures

Part C:    \mathbf{Q =220  \ J} to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = C_v=\dfrac{3}{2}R

For Part A:

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K

Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)

Q=93.5325 \ J

\mathbf{Q =94 \ J} to two significant figures

Part B. For hot temperature, C_v=\dfrac{5}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K

Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)

Q=155.8875 \ J

\mathbf{Q =160  \ J} to two significant figures

Part C. For an extremely hot temperature, C_v=\dfrac{7}{2}R

Q = mC_v\Delta T

Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K

Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)

Q=218.2425 \ J

\mathbf{Q =220  \ J} to two significant figures

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Explanation:

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