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
Weow that’s cool but what is your question
Answer:
1) 3.07kgm/s
2) 5.56kgm/s
3) 76.16N
4) 4.33kgm/s
5) 0.57s
6) -8.66J
Explanation:
Given
m = 0.221kg
v = 13.9m/s
θ = 25°
t = 0.073s
1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,
P(i) = mv(i)
P(i) = 0.221 * 13.9
P(i) = 3.07kgm/s
2) Magnitude of the change in momentum of the ball,
P(i,x) = P(i) cos θ
P(i,x) = 3.07 * cos25
P(i,x) = 3.07 * 0.9063
P(i,x) = 2.78
ΔP = 2P(i,x)
ΔP = 2 * 2.78 = 5.56kgm/s
3) magnitude of the average force exerted by the wall,
F(ave) = ΔP/Δt
F(ave) = 5.56/0.073
F(ave) = 76.16N
4) ΔP(z) = mv(f) - mv(i)
ΔP(z) = 0.221*-7.8 - 0.221*11.8
ΔP(z) = -1.72 - 2.61
ΔP(z) = 4.33kgm/s
5) F(ave) = ΔP/Δt
Δt = ΔP/F(ave)
Δt = 4.33 / 76.16
Δt = 0.57s
6) KE(i) = 0.5mv(i)²
KE(f) = 0.5mv(f)²
ΔKE = 0.5m[v(f)² - v(i)²]
ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]
ΔKE = 0.1105 ( 60.84 - 139.24 )
ΔKE = 0.1105 * -78.4
ΔKE = -8.66J
Answer:
Total work done to lift barbell = 890 J
Explanation:
Given:
Weight of barbell (F) = 445 N
Height (Distance) = 2 meter
Find:
Total work done to lift barbell = ?
Computation:
⇒ Work = Force(F) × Distance
⇒ Total work done to lift barbell = Weight of barbell × Distance
⇒ Total work done to lift barbell = 445 N × 2 meter
⇒ Total work done to lift barbell = 890 J
The net force on the box is:
50 + 30 - 65
= 15 Newtons
This can be used in conjunction with
F = ma
to calculate the acceleration of the box if its mass is known.