Answer:
1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%
Explanation:
For the first part, which is speed just before the crash, we can use energy conservation
Initial. Highest point
Em₀ = U = mg y
Final. Low point just before the crash
Emf = K = ½ m v²
Em₀ = Emf
m g y = ½ m v²
v = √ 2 g y
Let's calculate
v = √ (2 9.8 0.05)
v = 0.99 m / s
1) the moment before the crash is
p₀ = m v
p₀ = 0.221 0.99
p₀ = 0.219 kg m / s
After the collision, the car's speed is zero, so its moment is zero.
p = 0
2) change of momentum
Δp = p - p₀
Δp = 0- 0.219
Δp = -0.219 kg m / s
3) the reason is
Δp / p = 1
In percentage form it is 100%
Answer:
1. It undergoes reflection. 2. It undergoes refraction. 3. It undergoes diffraction.
Explanation:
1. It undergoes reflection. This is because it bounces off surfaces when incident on them.
2. It undergoes refraction. This is because it changes direction when it passes from one medium to another
3. It undergoes diffraction. This is because it spreads out when it passes through doors and windows similar in dimension to the dimensions of its wavelength
Answer:
I got you.. i'm in middle school and had that same question.
Explanation:
Refer to the diagram shown below.
The vertical distance traveled is
s = 25 m
The initial vertical launch velocity is zero.
Therefore
s = (1/2)*g*t²
where g = 9.8 m/s²
t = the time of flight, s
That is,
0.5*9.8*t² = 25
t² = 25/4.9 = 5.102
t = 2.26 s
Answer: 2.26 s
Because the position depends on the amount of time that has passed.
