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Answer:
-2.79 m/s²
Explanation:
Given:
v₀ = 20 m/s
v = 11 m/s
Δx = 50 m
Find: a
v² = v₀² + 2aΔx
(11 m/s)² = (20 m/s)² + 2a (50 m)
a = -2.79 m/s²
Round as needed.
Answer:
1. W = F d = 20 N * 6 m = 120 J
2. F = W / d = 60 J / 2 m = 30 N
3. d = W / F = 350 J / 85 N = 4.12 m
4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts
5. W = P t = 500 W * 120 sec = 60,000 J
6. t = W / P = 550 J / 310 W = 1.77 sec
Answer:
181.48 N
Explanation:
Calculate the area :
Area = pi * r² ;
pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m
Area 1, A1 = 3.14 * 0.1² = 0.0314 m²
Area 2, A2 = 3.14 * 0.9² = 2.5434 m²
Force, F = mass * acceleration due to gravity
F2 = 1500 * 9.8 = 14700 N
Force 1 / Area 1 = Force 2 / Area 2
Force 1 = (Force 2 / Area 2), * Area 1
Force 1 = (14700 / 2.5434) * 0.0314
Force = 5779.6650 * 0.0314
= 181.48 N
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
