Define brown horse:
a₁ = 2.2 m/s², acceleration
v₁ = 18 m/s, top speed
Define the black horse:
a₂ = 1.7 m/s², accelerastopn
v₂ = 20 m/s, maximum speed
To travel 2000m, each horse accelerates to maximum speed, and cruises to the finish line.
For the brown horse:
The time to attain maximum speed is
t₁₁ = v₁/a₁ = 18/2.2 = 8.182 s
The distance travelled during the acceleration is
s = (1/2)*a*t² = 0.5*2.2*8.182² = 73.636 m
The time to travel the remaining distance is
t₁₂ = (2000 - 73.636)/18 = 107.02 s
The total time of travel is
t₁ = t₁₁ + t₁₂ = 8.182 + 107.02 = 115.2 s (approx)
For the black horse:
Time to attain maximum speed is
t₂₁ = 20/1.7 = 11.765 s
Distance traveled while accelerating is
0.5*(1.7 m/s²)*(11.765 s)² = 117.647 m
Time to travel the remaining distance is
t₂₂ = (2000 - 117.647)/20 = 94.118 s
The total time of travel is
t₂ = t₂₁ + t₂₂ = 11.765 + 94.118 = 105.9 s (approx)
Conclusion:
The black horse wins the race in about 106 s, while the brown horse takes about 115 s
Answer: The black horse wins.
The change in velocity is 10 m/s in 5 seconds.
sinces a = change in velocity/time
a= 10/5
a=2
Answer:
Answer is 14.65
Refer below.
Explanation:
Refer to the picture for brief explanation.
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm