D = 110 m, t = 5 s
v o = 110 cs : 5 m = 22 m/s
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v = v o - a t
v = 0 m/s, v o = 22 m/s, t = 4 s
0 = 22 - 4 a
4 a = 22
a = 22 : 4
a = 5.5 m/s²
g = 9.80 m/s²
9.80 : 5.5 = 0.56
Answer:
The magnitude of its acceleration is 5.5 m/s or 0.56 g.
When the frequency decreases the wavelength is further apart. When it increases its closer together. Think about a flat line when the frequency is low the wavelengths are wider. When its a high frequency the squiggly lines on the moniter are taller and thinner so the wavelengths are not as wide and not that far from each other depending on how high the frequency is.
Answer:
Distance = displacement = 35m
Explanation:
The distance of the student is how far he has gone.
Distance = 25m + 10m
Distance = 35m
Displacement is the distance specified in specific direction. Since the student walk in the sane direction, thence the displacement is also 35m
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
