<span>Q = mL
</span>
We need to know the latent heat of fusion, L, for 0 degrees.
L = 336 kJ/kg, m = 2kg
where L is the latent heat for vaporization
Q = 2 * 336 = 672 kJ
For conversion between 0 and 100 degree Celsius
Q = mcθ, specific heat capacity = 4.2 kJ/kgK
Q = 2*4.2* (100 - 0) = 840 kJ
For conversion to steam at 100 degrees Celsius
<span>Q = mL , L = latent heat of vaporization = 2256 kj/kg
</span>
Q = 2 * 2256 = 4512 kJ
Total heat = 672 + 840 + 4512 = 6024 kJ
= 6 024 000 J
But 1 calorie = 4.2 J
Therefore 6 024 000 J will be: 6 024 000/4.2 ≈ 1 434 286 Calories
≈ 1 434 kCalories
Answer:
Using the range formula R = v^2 sin 2 theta / g
or v^2 = R * g / sin 86.4
v^2 = 3.14 m * 9.81 m/s2 / .998
v^2 = 30.9 m^2 / s^2
v = 5.56 m/s
This hasn't really proved the question - this would give
vy = 5.56 * sin 43.2 = 3.81 m/s
vx = 5.56 * cos 43.2. = 4.05 m/s
t = 1.57 / 4.05 = .387 sec to reach the waterfall
h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls
There seems another way to do this
vy / vx = tan 43.2 vy = .939 vx
h = vy t - 1/2 g t^2 and t = 1.57 / vx
h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2
Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s
Answer:
Why do I keep getting spam calls? Experts credit the ascendance of spam phone calls to fundamental problems with caller ID, a phone system where anyone can operate as a carrier, the inability to detect bad callers, and a number of bad actors exploiting those flaws to drive billions of calls to American phones.
Explanation:
Answer:
huh? do you need help on math?
Explanation:
what do you mean?
We actually cant, really. Black Holes are really very powerful and don't just happen in some random place in the cosmos. We have got shots of Black holes that you cannot even see, but all the other big black holes that look like something straight of a sci-fi movie that look real, are really just photo- shopped. good question. Hope This Helped.
