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How long would it take you to swim across a lake that is 900 meters across if you swim at 1.0 ms2?
1 m/s² (1 metre per second per second) is a measure of acceleration, NOT speed. Assuming you mean a swim rate of 1.0 m/s (1 metre per second), then: time = distance ÷ speed = 900 m ÷ 1 m/s = 900 s 60 s = 1 min → 900 s = 900 ÷ 60 min = 15 minutes ---------------------------------- Assuming you really do mean that you swim with a constant acceleration of 1 m/s² and you start at 0 m/s, then: s = ut + ½at² s = 900 m u = 0 a = 1 m/s² t = unknown → 900 m = 0 m/s × t + ½ × 1 m/s² × t² → 900 m = ½t² → t² = 1800 s² → t ≈ 42.43 s If you start at 0 m/s and accelerate at 1 m/s², then it will take approx 42.43 seconds to cover 900 m.
Why do units of acceleration have two time units?
Acceleration = (v - u)/t Replace each variable in the equation with their units Acceleration = (m/s - m/s) / s Same common denominator (m - m/s)/s Apply divison of fractions (m-m)/s*s (m/s^2) Note: I know m - m = 0 but they are just units expressing metres!
When the bus decelerates uniformly at 5 point 3 m per s2 it slows from 8.5 m per s to 0 m per s what is the time interval of acceleration for the bus?
Vf = V0 + at --> 0 = (8.5 m/s) - (5.3 m/s²)t -> t = (8.5 m/s)/(5.3 m/s²) = 1.60377 s
What is the average speed of an object whose distance-time graph indicates that it travels 2 m in 2 s and then travels another 80 m during the next 40 s?
total distance = 2m + 80m = 82 m total time = 2s + 40s = 42s → average speed = total distance / total time = 82m / 42s = 1 20/21 m/s ≈ 1.95 m/s
What is the Z score equation?
Given a random variable X with mean M and standard deviation S, Z = (X - M)/S
