Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Answer: (1) The correct answer is A.
(2) The correct answer is D.
Explanation:
(1)
Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.
Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.
Therefore, the correct answer is A.
(2)
Amplitude is the maximum displacement in the medium from the rest position.
The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.
Therefore, the correct option is D.
Answer:this is confusing and what subject is this
Explanation:
Answer:
Explanation:
Given
See attachment for the graph
Required
Determine the frequency
Frequency (F) is calculated as:
Where
T = Time to complete a period
From the attachment, the wave complete a cycle or period in 3 seconds..
So:
--- Approximated
Answer:
a) A = [m /s²]
, B = [m / 
]
, b) v = a to + 2B/3
Explanation:
a) to define the units of the constants, we see that the left part has units of acceleration so the right part must also have the same units
A = [m /s²]
B = [m / s²]
B = [m / ]
b) We must use the definition of acceleration
a = dv / dt
dv = adt
We integrate
∫ dv = ∫ (A + B 
) dt
v = a t + B 2/3
We evaluate between:
the lower limit t = 0 v = o and the upper limit t = to v = v
v = a to + 2B/3
c) Let's use the definition of speed
v = dx / dt
dx = v dt
∫ dx = ∫ (a t + 2B/3 ) dt
x = a t²/2 + 2b/3 2/5 
For the interval t = 0 and t = to, we evaluate the integral
x = A/2 t₀² + 4B/15 √t₀⁵
