How to convert km/h to m/s squared

Thread starter mike98020205
Start date Sep 25, 2005
Tags

Convert

AI Thread Summary

To convert kilometers per hour (km/h) to meters per second (m/s), it's important to recognize that the conversion involves time and distance, not acceleration. The formula involves multiplying by the conversion factors: 1 km equals 1000 meters and 1 hour equals 3600 seconds. The confusion arises from the mention of "squared," which is not applicable in this conversion. The correct approach is to use the formula: (km/h) * (1000 m / 1 km) * (1 hr / 3600 s) to arrive at m/s. This method clarifies the conversion process effectively.

Sep 25, 2005

mike98020205

I guess I'm confused on how to convert km/h to m/s squared.

I understand 1min/60s=360s/1 hr but this seems different. the seconds squared ared throwing me off. I know there's a 1000 meters in a kilometer and 360 seconds in an hour. I guess I don't know how to apply these numbers. Thanks for your help.

Physics news on Phys.org

Sep 25, 2005

Poncho

1hr = 3600s

The seconds aren't squared

<br /> \frac{km}{hr}*\frac{hr}{3600s}*\frac{1000m}{1km}=\frac{m}{s}<br />

Sep 25, 2005

mike98020205

Ty so much.

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »