The answer is number (3) if am wrong am sry
Answer:
The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
Explanation:
To calculate the final pressure 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,

where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:

Putting values in above equation, we get:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
Answer:
B) The lightbulb uses 4,104,000 J more than the stereo.
Answer:

Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

Where
is the length of the string and
the velocity of propagation. Use this expression to find the value of
.

The velocity of propagation is given by the expression:

Where
is the desirable variable of the problem, the linear mass density, and
is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

With the value of the tension and the velocity you can find the mass density:

