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:
5 meters per second
Explanation:
5m is the distance
5m west is the vector
5m per second is the velocity
5m per second west is unknown
Answer:
so rate constant is 4.00 x 10^-4 
Explanation:
Given data
first-order reactions
85% of a sample
changes to propene t = 79.0 min
to find out
rate constant
solution
we know that
first order reaction are
ln [A]/[A]0 = -kt
here [A]0 = 1 and (85%) = 0.85 has change to propene
so that [A] = 1 - 0.85 = 0.15.
that why
[A] / [A]0= 0.15 / 1
[A] / [A]0 = 0.15
here t = (79) × (60s/min) = 4740 s
so
k = - {ln[A]/[A]0} / t
k = -ln 0.15 / 4740
k = 4.00 x 10^-4
so rate constant is 4.00 x 10^-4
Answer:Newton's three laws of motion relate to each other in that they lay a foundation for the principles of things in motion, then build upon that foundation. For example, the first law of motion,...
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