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plz help me i am getting incorrect answer
plz show working as well

3. A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.5 s later. How high was the cliff? How far from its base did the diver hit the water?
my answer is height of cliff = 50.23 and hits 5.76m away from base

7. Suppose the space shuttle is in orbit 416 km from the Earth's surface, and circles the Earth about once every 92.8 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth's surface.
my answer 0.00540gm/s^2

8. Because the Earth rotates once per day, the effective acceleration of gravity at the equator is slightly less than it would be if the Earth didn't rotate. Estimate the magnitude of this effect, given that the radius of the Earth is 6.38 106 and that, at the equator, it rotates at 464 m/s. What percentage of g is this?
my answer 0.344%
plz help me

2007-09-23 04:41:50 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

3.
h=0.5 g t^2
h= 0.5 x 9,81 (3.5)^2 = 60.1 m

Horizontal velocity remains constant
S= vt= 1.8 x 3.5= 6.3 m

7.
a=(V^2)/R
R= H+r=416 E+3 + 6.38 E+6=6.80 E+6
V=2 pi R/T
a=[(2 pi R/T)^2]/R
a= 4R (pi/T)^2
a=4 x 6.80 E+6(3.14... / 92.8 x 60)^2 = 8.66
a=8.66 or
a=0.88g
(It better pick up some speed or it will fall back to Earth)

8.
again
a=(V^2)/r
g-effective= g - a

g-effective= 9.81 - (464)^2/6.38 E+6=
g-effective= 9.81 - 0.0339=9.78 m/s^2

2007-09-23 04:55:52 · answer #1 · answered by Edward 7 · 0 0

please show your solutions so I can know why your answers are incorrect.

2007-09-23 11:56:31 · answer #2 · answered by zsm28 5 · 0 0

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