A plumb bob does not hang exactly along a line directed to the center of the Earth's rotation. How much does the plumb bob deviate from a radial line at 32.5° north latitude? Assume that the Earth is spherical.
answer is wanted in degrees
2007-02-16 05:36:25 · 3 answers · asked by sabresfan58 1 in Science & Mathematics ➔ Physics
The acceleration from the earth's rotation at 32.5° lat is
a = Re*cos32.5°*ω² = 2.84E-5 m/s²
Combining that with g at Lat 32.5 gives a deviation of roughly 8.93E-5°. The deviation will be toward the equator.
Draw a diagram with 9.8 m/s² acting toward the center and the value of a above acting outward perpendicular to the axis of the earth. This will give you a triangle with 2 sides and one angle known.
The tilt of the earth's axis doesn't enter into the answer.....
2007-02-16 07:07:36 · answer #1 · answered by Steve 7 · 1⤊ 0⤋
The deal is, that air drag is usually proportional to the sq. of speed. that's genuine for any type of drag this is ruled via impression rigidity (think of air and water at severe speeds). that's no longer genuine for drag this is ruled via viscosity (think of oils at sluggish speeds). At terminal speed, the drag rigidity is thru definition equivalent to the burden of the falling merchandise. this is the only way that the article can fall at a persevering with speed, for this reason the call. Weight is m*g Drag at terminal speed: D_term = m*g via the character of the rigidity of drag, that's proportional to the sq. of speed. Make a pretend formulation for drag, with a premier consistent (ok) in front of speed squared. we are able to connect contraptions to it, of Newton-seconds^2/meter^2, in case you elect. we are able to grant ok a concept, yet this is a topic rely for yet another day. D = ok*v^2 we've an interest interior the case, whilst D = D_term/2. And we would desire to be conscious of what speed that occurs at. Given: D_term = ok*v_term^2 build the case of: D_term/2 = ok*v^2 remedy first equation for ok: ok = D_term/v_term^2 remedy 2nd equation for v: v = sqrt(D_term/(2*ok)) Plug in ok: v = sqrt(D_term/(2*(D_term/v_term^2))) Simplify: v = v_term/sqrt(2) Given v_term = 60 m/s^2 answer: v = 40 two m/s
2016-12-17 11:28:22 · answer #2 · answered by moncalieri 4 · 0⤊ 0⤋
Did you take the earth's 23 degree axis into consideration? Perhaps this is a stupid question.
2007-02-16 06:13:53 · answer #3 · answered by Tom H 4 · 0⤊ 2⤋