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Q: What does m s time mean?
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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

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What does the sexual term s and m mean?

S=Sado M=Masochism = sadomasochism


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180 IS THE M S WITH 3 D


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E= Early T= Time D= Dissmissal M= Monthly T= Time Y= Yesterday S=Study


A dog is running at 6.45 m/s. His human calls to him and 3.19 seconds later he is running at 6.56 m/s. What is his acceleration during this time?

0.034 m/s/s


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.


Car is traveling at a constant speed of 12 m s when the driver accelarates the car reaches a speed of 26 m s in 6 s what is the average accelaration of the car?

The change in velocity is 26 m/s - 12 m/s = 14 m/s. The time taken for this change is 6 seconds. Therefore, the average acceleration is (change in velocity / time) = 14 m/s / 6 s = 2.33 m/s^2.


A runner covers the last straight stretch of a race in 4 s during that time he speeds up 5 m s to 9 m s what is the runbers acceleration in this part of the race?

Vi = 5 m/s A = (V_f- V_i )/t = (9 m⁄s- 5 m⁄s )/(4 sec) = (4 m⁄s )/(4 sec) = Vf = 9 m/s Change in time = 4 sec


What does s m a r t mean?

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What has the author S M James written?

S. M. James has written: 'Time in primary mathematics'


How do you calculate standard normal variate?

You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).You use the z-transformation.For any variable X, with mean m and standard error s,Z = (X - m)/s is distributed as N(0, 1).