Answer:
Tarzan will be moving at 7.4 m/s.
Explanation:
From the question given above, the following data were obtained:
Height (h) of cliff = 2.8 m
Initial velocity (u) = 0 m/s
Final velocity (v) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 2.8)
v² = 0 + 54.88
v² = 54.88
Take the square root of both side
v = √54.88
v = 7.4 m/s
Therefore, Tarzan will be moving at 7.4 m/s at the bottom.
The difference between delta and harbor is is that delta is the fourth letter of the modern greek alphabet while harbor is a sheltered expanse of water, adjacent to land, in which ships may dock or anchor, especially for loading and unloading.
Answer:
alpha=53.56rad/s
a=5784rad/s^2
Explanation:
First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)
Now, we can calculate the angular acceleration (w0=0rad/s)
with this value we can compute the angular velocity
and the tangential velocity of point B, and then the acceleration of point B:
hope this helps!!
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
