Answer:
6 m/s
Explanation:
12m / 2s = 6 m/s
Hope that's the answer you seek.
Answer:
=118.8 K= 154.2°C
Explanation:
COP_max of carnot heat pump= 
where T_H and T_C are temperatures of hot and cold reservoirs
Also COP=
in the question 
⇒
heat is added directly to be as efficient as via heat pump

and T_H= 24° C= 297 K

on calculating the above equation we get
=118.8 K
the outdoor temperature for efficient addition of heat to interior of home
=118.8 K= 154.2°C
That’s really easy ask your teacher and also peace happy
Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.