Recent content by san512

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    Why am I getting the wrong speed for a train using the Doppler effect formula?

    A) 7.3 m/s B) 12 m/s C) 21 m/s D) 32 m/s E) 42 m/s I get 9.4 m/s
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    Why am I getting the wrong speed for a train using the Doppler effect formula?

    Homework Statement A train moving at a constant speed is passing a stationary observer on a platform. A flute player is playing a note with a frequency of 940Hz. after the flute has passed, the observer hears the sound frequency of 915Hz. What is the speed of the train? The speed of sound in...
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    Car tire kinetic energy at end of 0.25 mile acceleration

    Got it. Thanks a lot. You helped me out a lot.
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    Need help with physics? Welcome to PF!

    I.m a college student struggling through physics and i find this website very helpful!
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    Car tire kinetic energy at end of 0.25 mile acceleration

    i got it. thanks a lot. if you don't mind could you help me with applying significant figures? i have trouble with problems that need a lot conversions. so for example, 12 miles to m. i see 2 in miles so would that mean i have to round m to 2 significant figures as well?
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    Car tire kinetic energy at end of 0.25 mile acceleration

    but the question is asking for rotational KE. don't i need to calculate w for that?
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    Car tire kinetic energy at end of 0.25 mile acceleration

    okay. so using vf= 2d-vi/t 2*402.25m/15s= 53.63m/s relating linear motion with rotational motion v/r=w 53.63m/s/.23m =233.17 is that right?
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    Car tire kinetic energy at end of 0.25 mile acceleration

    how would i find angular velocity and what units am i missing?
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    Car tire kinetic energy at end of 0.25 mile acceleration

    Homework Statement The average car today has a mass of 1100 kg, and when accelerating from rest, covers 0.25 miles in 15 seconds. Each rim and tire together has a diameter of 46 cm and a mass of 9.1kg. If we agree the rim and tire have the shape of a solid disk that rotates through its...