Answer:
The voltage of the battery
The specific heat of water is 4.186.
Answer:
308,000 or 30.8×10^3
Explanation:
v=f×lamda
v is ?, f is 875Hz, lamda is 352m
v=875×352
v=308,000
v=30.8×10^3 m/s
Answer:
Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C. For comparison sake, it only takes 385 Joules of heat to raise 1 kilogram of copper 1°C.
Explanation:
Answer:
1/5 km/min
Explanation:
the formula for velocity is distance/time
so if i plug in the distance and time i get 5/25 or 1/5
Hope this helps!