In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g=

Where 1 kg=1000 g

To determine the amplitude of oscillations.

Energy for oscillator=

Where =Angular frequency

A=Amplitude

Using the formula

Therefore, the amplitude of oscillation=A=0.199

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

a.

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

b.

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

Explanation:

= initial resistance

= temperature coefficient of resistivity

= final temperature    

= initial temperature

= 3 ohm

= 0.0045

        R =

        36 =

Thus, it can be concluded that the temperature of the light bulb at 12.0 V is 7626.33 K.