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Does anyone want to help me with my physics????

Early skeptics of the idea of a rotating Earth said that the fast spin of Earth would throw people at the equator into space. The radius of Earth is about 6400 km. Show why this objection is wrong by calculating:

A) the speed of a 97 kg man at the equator; The force acting on him to accelerate him in a circle around the Earth

C) The weight of the man;

D) How much force is holding him to the surface of the Earth

I'm so confused... much thanks for any help Share on other sites

A) v = 2*pi*R/T

R = Radius of Earth, 6400 km.

T: T is the period of rotation, i.e. one day. One day = 24 hours. I don't think it's necessary to convert from hours to seconds, since km/h seems like a good unit to stick with.

v = 1675.52 km/h

Now you have v. F = ma.

a = centripetal acceleration = v^2/r

a = 438.649 km/h^2

F = 42549 km*kg/h^2 (or 3.28 N (kg*m/s^2), if you prefer)

C) Weight = mg. Since gravity is 9.81 meter/second^2, convert this to km/h^2. (g = 127138 km/h^2)

W = 1.233 * 10^7 km*kg/h^2 (or 951.57 N)

D) W - F = a positive number. Therefore, I am at my computer, and not hurtling towards the atmosphere, about to be burnt up.

Oh, and Back_Lit, though Magewout's solution is correct, when finding C, make sure you find the specific heat (J/g*Celsius), and not the heat capacity (J/Celsius).

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Don't know if I'm totally correct...

1) I'd say the speed of a man at the equator would be (2 . pi . 6400 km) / 24h , seeing as the earth turns around in 24 hours.

2) This is a long time ago... with the speed you've just calculated, the radius, and his weight, you would be able to calculate that force

3) The weight of the man would be 97 kg . 9.81 N/kg

4) The force holding him to the surface would be gravity, so Fz = m . g . h = 97 kg . 9.81 N/kg . 6400 km

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*woot first post *

The m needed would be the marble's weight, 0.250 kg. Also, the you can assume that if the water keeps boiling, the final temperature of the marble will be 100 degrees C.

Q = m[marble] . c[marble] . (100 - T[marble])

m[marble] = 0.250 kg

c[marble] would be = c[glass] (look that one up)

T[marble] = -23 C Now I can finish my damn HW for physics AND get 8 hrs sleep. Thanks a lot!

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oh good thanks. i had figured out part A and then i was lost on all of the rest Share on other sites

I was wondering whether you can assume the final temperature would be 100 degrees C...should've posted my thought on that. That's a poor question IMO though.

As for the rotating Earth problem:

A) Note that the period of rotation = 24 hours = 24 * 60 seconds. use v = Circumference of the Earth / T (circumference = 2 * pi * r) Use that the centripetal force = v^2/r

C) W = m * g

D) W - m * v^2/r

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We did it.

Care to elaborate?

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can someone tell me how to put pictures in my signature? The copy paste feature doesn't seem to work on this level and thats the only thing that i can think of.

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can someone tell me how to put pictures in my signature? The copy paste feature doesn't seem to work on this level and thats the only thing that i can think of.

http://www.ocremix.org/phpBB2/viewtopic.php?t=11654&start=0

I would explain it myself, but that should make it easier.

If you want a sig made, ask here.

Hopefully that answers your question, just post in the help/newbies section next time instead of a random thread EDIT: I like the sig.

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We did it.

Care to elaborate?

Yes plz I just read this thread and now I feel entitled to see some results.

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We did it.

Care to elaborate?

Yes plz I just read this thread and now I feel entitled to see some results.

Oh sorry. If I remember I'll take webcam pics of it tomorrow morning.

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