Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof
1 answer:
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Answer:
r = 1.61 x 10^{11} m
Explanation:
energy radiated (H) = 2.7 x 10^31 W
surface temperature (T) = 11,000 k
assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}
we can find the radius of the star from the equation below
H = A x ε x σ x T^{4}
where area (A) = 4 x π x r^{2} (assuming it is a sphere)
therefore the equation becomes
H = 4 x π x r^{2} x ε x σ x T^{4}
2.7 x 10^31 = 4 x π x r^{2} x 1 x 5.67 x 10^{-8} x (11,000)^{4}
r =
r = 1.61 x 10^{11} m
Answer:
Explanation:
For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:
v₀ = 48 ft/s
a = -32 ft/s/s
v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:
v² = v₀² + 2aΔx and filling in:
Δx and
0 = 2300 - 64Δx and
-2300 = -64Δs so
Δx = 36 feet.
Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:
v = v₀ + at and filling in:
0 = 48 - 32t and
-48 = -32t so
t = 1.5 sec. That's part a. Onto part b:
The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:
Δx = v₀t + and filling in:
and
and
0 = 16t(3 - t) so
t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:
We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:
and get everything on one side and factor it again:
and we find that
1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.