This question is incomplete the complete question is
A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?
Answer:
(a) Xs=0.459m
(b) t=0.984 s
(c) Vc=6.65 m/s
Explanation:
(a) To reach maximum distance
(b) For Time
To find t we must find t1 and t2
as
t=t1+t2
For T1
For T2
For Total Time
t=t1+t2
t=0.306+0.6789
t=0.984s
(c) To find Vc
Vc=Vb+gt2
Vc=(0)+(9.8)(0.6789)
Vc=6.65 m/s
Work can be defined as the energy transferred from a body to its sorroundings, the energy spent to move a body, or the energy you need to alter a charged particle, so no energy, no work; thus, the statement is true.
R is proportional to the length of the wire:
R ∝ length
R is also proportional to the inverse square of the diameter:
R ∝ 1/diameter²
The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.
Calculate the scale factor due to the changed length:
k₁ = 2800/2700 = 1.037
Scale factor due to changed diameter:
k₂ = 1/(0.1/0.26)² = 6.76
Multiply the original resistance by these factors to get the new resistance:
R = R₀k₁k₂
R₀ = 9850Ω, k₁ = 1.037, k₂ = 6.76
R = 9850(1.037)(6.76)
R = 69049.682Ω
Round to the nearest hundredth:
R = 69049.68Ω
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
Calculate velocity at halfway to the ground.
vfinal = root 2ad
v = root (2*9.81m/s^2*25)
v = 69.367175234 m/s
Kinetic energy = 1/2mv^2
Kinetic energy = 1/2 * (69.367175234 m/s^s^2
Ek = 2405.9025 Joules