You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.

Answer:

69.69 g

Explanation:

Evaporation of water will take out latent heat of vaporization.  Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg

Heat taken up by evaporating water

= 2260000 x m J

Heat lost by body

= mass x specific heat of body x drop in temperature

60 x 3500 x .750  ( specific heat of human body is 3.5 kJ/kg.k)

= 157500 J

Heat loss = heat gain

2260000 m= 157500

m = .06969 kg

= 69.69 g

Final answer:

Approximately 78 grams of water would need to evaporate from a 60.0-kg person to lower their body temperature by 0.750ºC. This calculation is based on the principles of thermodynamic heat transfer and the specific body temperature, latent heat of water vaporization, and specific heat capacity of the human body.

Explanation:

To calculate the amount of water mass from an individual's body that would need to evaporate to reduce their body temperature, we can use the principle of thermodynamic heat transfer. The basic equation is Q = mLv, where Q is the heat absorbed or lost, m is the mass, and Lv is the latent heat of vaporization.

In this case, knowing that at body temperature of 37.0°C, the latent heat of water vaporization (Lv) is approximately 2430 kJ/kg, we substitute these numbers. Given our desire to reduce body temperature by 0.750°C in a 60 kg human, we first calculate the amount of heat to dissipate (Q) using Q = mcΔT, where c is the specific heat capacity of the human body (roughly equivalent to that of water, 4.184 kJ/kg°C), m is the mass, and ΔT is the change in temperature.

The calculation is as follows:

Q = (60 kg)(4.184 kJ/kg°C)(0.750°C) = ~189 kJ

Next, we substitute Q into the Q = mLv equation to determine the mass m:

m = Q / Lv = 189 kJ / 2430 kJ/kg = 0.078 kg, or 78 grams

Hence, around 78 grams of water would need to evaporate from a 60.0-kg person to lower their body temperature by 0.750ºC.

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When two bodies come into close touch with one another, a collision occurs. In this instance, the two bodies quickly exert forces on one another. The collision changes the energy and momentum of the bodies that are interacting.

Briefing

the system's initial kinetic energy, KEi, is equal to 0.5 * 4 * 1.8 2 plus 0.5 * 6 * 0.2 2 J.

KEi = 6.6 J

The system's ultimate kinetic energy, KEf

, following the collision is equal to 0.5 * 4 * 0.6 + 0.5 * 6 * 1.4 J.

KEf = 6.6 J

since KEi = KEf

Perfectly elastic is the collision

the appropriate response is A) completely elastic.

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Answer:

option the correct is B

Explanation:

Let's analyze the different options, for a closed system

- an internal reaction changes the system, but does not affect the surrounding environment

- Heat, is a means of transfer that occurs when two bodies are in contact, one of the body can be a closed system since the only thing that happens is thermal transfer, without movement of the system itself. This is the correct result.

- Work implies a movement whereby the system must be mobile, it is not an option

- Pressure change. change in the system, but does not affect the environment

- Mass transfer is not possible in a closed system

After analyzing each option the correct one in B

The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t

Where,

θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration =  4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

Hence the angular velocity will be 22.41 rad/s.

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Answer:

= 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

where is the final angular velocity, the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

Finally, solving for :

= 22.41rad/s