Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
Answer:
I don’t understand Espanol
Explanation:
sorry
All the bats living in a cave form a __colony__of bats in the cave ecosystem.
example force. because you can say" I applied 3 newtons downward to the floor".
force has magnitude and direcion
