Answer:
3.982 kg
Explanation:
The latent heat of vaporization = 540 cal/g
= 5.4 ×10⁵ cal/kg
L = 5.4 ×10⁵ × 4.19
= 2.26 × 10⁶ J/kg
Q = 12 × 750,000
= 9, 000, 000
= 9 × 10⁶ J
the maximum number of kg of water (at 100 degrees Celsius) that could be boiled into steam (at 100 degrees Celsius) is:
= 
= 3.982 kg
Answer:
6 meters away
Explanation:
6*1.4= 8.4 which is pretty close
I don't think so. How are you supposed to write the answer?
Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.
a) = (1.82i - 9.69k)m/s2
b) magnitude = 9.85m/s2
c) direction = 280.64 degree
Explanation:
The detailed and step is shown in the attachment.
