T = 2*V₀*sin (α) / g
t = 2*1770*sin (26.3°) / 9.8 ≈ 2*1770*0.443 / 9.8 ≈ 160 s or 2.7 min
Hello, love! The answer is True, or T, on Edge2020.
Hope this helped!
~ V.
I'm assuming your question refers to the train travels 81 km in 2 hours, as in that's the total distance covered versus the speed for those 2 hours, whereas the 90 km in 2 hours the second time around is the total distance not speed I would assume.
Now... if it was going 81 km for the first 2 hours and 90 km for the second 2 hours then the average speed would be the mean of these numbers, with that being 85.5 km. Though, I doubt that's your question.
With that said, 81 km covered by 2 hours and 90 km covered by 2 more hours. To acquire the km an hour average, we'll have to divide the distance by how many hours it traveled:
81 / 2 = 40.5
90 / 2 = 45
Meaning the train was going 40.5 km an hour for the first 2 hours and 45 km an hour for the second 2 hours. Now, to find the mean:
40.5 + 45 = 85.5 / 2 = 42.75
In-case you were wondering, the mean is the sum of all the numbers in a set, divided by the total amount of numbers in that set, take for instance:
3, 5, 9, 2, 1, 5. -> There are 6 numbers in this set.
3 + 5 + 9 + 2 + 1 + 5 = 25 -> 25 is the sum of these numbers.
25 / 6 = 4.2 (estimated) -> 4.2 is roughly the mean or average between the original set.
Anyhow, with that aside the average speed between this I would believe would be 42.75 km an hour. I hope that helps, have a great rest of your day! ^ ^
| | Ghostgate (Alter) | |
Velocity depends on the straight-line distance between your start-point and your end-point, regardless of what route you follow to get there.
If you stop at the same point where you started, then that distance is zero, no matter how far you drove before you returned to your start-point.
So the average velocity around any "CLOSED" path is <em>zero. (A)</em>
Answer:
3.6 × 10⁵ N/C = 360 kN/C
Explanation:
Let R = 2.0 cm be the radius of the sphere and q = -8.0 nC be the charge in it. Let q₁ be the charge at radius r = 1.0 cm. Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πr³
q₁ = q(r/R)³. The electric field due to q₁ at r is E₁ = kq₁/r² = kq(r/R)³/r² = kqr/R³
The electric field due to the point charge q₂ = 5.0 nC is E₂ = kq₂/r².
So, the magnitude of the total electric field at r = 1.0 cm is
E = E₁ + E₂ = kqr/R³ + kq₂/r² = k(qr/R³ + q₂/r²)
E = 9 × 10⁹(-8 × 10⁻⁹ C × 1 × 10⁻² m/(2 × 10⁻² m)³ + 5 × 10⁻⁹ C/(1 × 10⁻² m)²)
E = 9 × 10⁹(-1 × 10⁻⁵ + 5 × 10⁻⁵)
E = 9 × 10⁹(4 × 10⁻⁵)
E = 36 × 10⁴ N/C = 3.6 × 10⁵ N/C = 360 kN/C