Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s
F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)
F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
<span>F = -47,222 N The negative sign means that the force vector is </span>
<span>applied AGAINST the momentum vector of the rhinoceros.</span>
Answer:
1) a rubber band
2) the spring of retractable pen
3) a spring loaded toy gun
Explanation:
Hooke's law states that; provided the elastic limit of a material is not exceeded, the force exerted on an elastic material is directly proportional to its extension. This relationship was first captured by Robert Hooke in 1660 when he asserted that 'as the extension, so is the force!'.
Hooke's law generally deals with elastic or stretchable materials. These materials can be deformed, but returned to their original shapes when the deforming force is removed. This deforming force causes an extension in the material which is directly proportional to the deforming force. That is F= Kx where K is the called the force constant, F is the deforming force and x is the magnitude of extension brought about by the force.
Various real life applications of Hooke's law have been listed in the answer. Any material that makes use of a loaded spiral spring or indeed any kind of elastic material obeys Hooke's law.
Answer:
70 kg is the mass of the object
Explanation:
This question can be solved with this simple formula:
Weight force = mass . gravity
686 N = mass . 9.8 m/s²
686 N / 9.8 m/s² = mass → 70 kg
Note → 1N = 1 kg . m / s²
Answer:
98.15 lb
Explanation:
weight of plane (W) = 5,000 lb
velocity (v) = 200 m/h =200 x 88/60 = 293.3 ft/s
wing area (A) = 200 ft^{2}
aspect ratio (AR) = 8.5
Oswald efficiency factor (E) = 0.93
density of air (ρ) = 1.225 kg/m^{3} = 0.002377 slugs/ft^{3}
Drag = 0.5 x ρ x 
x A x Cd
we need to get the drag coefficient (Cd) before we can solve for the drag
Drag coefficient (Cd) = induced drag coefficient (Cdi) + drag coefficient at zero lift (Cdo)
where
- induced drag coefficient (Cdi) =
(take note that π is shown as n and ρ is shown as 
)
where lift coefficient (Cl)= 
=
= 0.245
therefore
induced drag coefficient (Cdi) = = 
= 0.0024
- since the airplane flies at maximum L/D ratio, minimum lift is required and hence induced drag coefficient (Cdi) = drag coefficient at zero lift (Cdo)
- Cd = 0.0024 + 0.0024 = 0.0048
Now that we have the coefficient of drag (Cd) we can substitute it into the formula for drag.
Drag = 0.5 x ρ x x A x Cd
Drag = 0.5 x 0.002377 x (293.3 x 293.3) x 200 x 0.0048 = 98.15 lb
