'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.
I think there's a typo because the answer I'm getting is very large.
This is what I'm getting
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c = speed of light
c = 3.0 x 10^8 m/sec approximately
This is roughly 300 million meters per second
The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.
distance = rate*time
d = r*t
d = (3.0 x 10^8) * (7.0 x 10^2)
d = (3.0*7.0) x (10^8*10^2)
d = 21.0 x 10^(8+2)
d = 21.0 x 10^10
d = (2.1 x 10^1) * 10^10
d = 2.1 x (10^1*10^10)
d = 2.1 x 10^11 meters
d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)
Answer:
2.55 × 10³ J =2.55 kJ
Explanation:
Specific heat capacity of ice = 37.8 J / mol °C
Specific heat capacity of water = 76.0 J/ mol °C
Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁
Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .
Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .
Total heat = Q = Q₁ + Q₂ + Q₃
Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j
Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j
Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j
Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j
= 2547.039 j = 2.55 × 10³ J =2.55 kJ
